The Ceres is (a).The largest asteroid identified to date is correct option.
With a diameter of around 590 miles (940 km), Ceres is the biggest asteroid between Mars and Jupiter. It is regarded as a minor planet, like Pluto, and was found in 1801 by Giuseppe Piazzi. Ceres is made of rock and ice, and due to its low density, it is likely that it contains a substantial amount of water ice inside.
The NASA Dawn spacecraft visited Ceres in 2015 and orbited the dwarf planet for more than three years, gathering useful information and taking stunning pictures of it.
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A flat uniform circular disk (radius = 5.44 m, mass = 150 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a frictionless axis perpendicular to the center of the disk. A 47.0-kg person, standing 1.54 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.80 m/s relative to the ground. Find the resulting angular speed (in rad/s) of the disk.
The resulting angular speed of the flat, uniform circular disk is 0.237 rad/s.
What is angular momentum?Angular momentum is a fundamental concept in physics that describes the rotational motion of an object around an axis. It is defined as the product of the moment of inertia of an object and its angular velocity with respect to a chosen axis.
We can use conservation of angular momentum to solve this problem. The initial angular momentum of the disk is zero because it is stationary. The final angular momentum of the system (disk + person) is:
L = Iω
where I is the moment of inertia of the disk and person about the axis of rotation, and ω is the resulting angular speed of the disk.
The moment of inertia of the disk about its axis is:
I_disk = (1/2)mr²
where the disk's radius is r and its mass is m. Substituting the given values, we get:
I_disk = (1/2)(150 kg)(5.44 m)² = 2226.24 kg·m²
The moment of inertia of the person about the axis can be approximated as:
I_person = mr²
where r is the distance from the axis to the person. Substituting the given values, we get:
I_person = (47.0 kg)(1.54 m)² = 109.64 kg·m²
The total moment of inertia of the system is:
I = I_disk + I_person = 2226.24 kg·m² + 109.64 kg·m² = 2335.88 kg·m²
The final angular momentum of the system is:
L = Iω
where ω is the resulting angular speed of the disk. Substituting the given values, we get:
(2335.88 kg·m²)ω = (197.64 kg·m²/s)(2.80 m/s)
Solving for ω, we get:
ω = (197.64 kg·m²/s)(2.80 m/s) / (2335.88 kg·m²) = 0.237 rad/s.
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What is the period, in seconds, of this mass?
The period (in second) of the 0.95 Kg mass, given that it has an amplitude of 0.21 m and an angular velocity of 9.5 rad/s is 0.66 second
How do i determine the period of the mass?From the question given above, the following data were obtained:
Mass (m) = 0.95 KgAmplitude (A) = 0.21 mAngular frequency (ω) = 9.5 rad/sPi (π) = 3.14Period (T) =?The period of the mass can be obtained as shown below:
ω = 2π/ T
9.5 = (2 × 3.14) / T
9.5 = 6.28 / T
Cross multiply
9.5 × T = 6.28
Divide both sides by 9.5
T = 6.28 / 9.5
T = 0.66 second
Thus, from the above calculation, we can conclude that the period of the mass is 0.66 second
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If the wave pictured above oscillates up and down 25 time in 10
seconds, what is the frequency?
Answer:
Frequency = 25 / 10 = 2.5 Hz
Explanation:
If the wave oscillates up and down 25 times in 10 seconds, then the frequency can be calculated as follows:
Frequency = Number of oscillations / Time
In this case, the number of oscillations is 25, and the time is 10 seconds. Therefore, the frequency is:
Frequency = 25 / 10 = 2.5 Hz
So the frequency of the wave is 2.5 Hertz, which means it completes 2.5 cycles or oscillations in one second.
On Earth the gravitational field strength is 10N/kg. What would be the weight of someone on Earth who had a mass of 75kg?
750N/kg
750N
75N
7500kg
The weight of someone on Earth with a mass of 75kg is 750N.
option B.
What would be the weight of someone on Earth?The weight of someone on Earth is calculated using Newton's second law of motion.
W = mg
where;
m is the massg is acceleration due to gravityFor the mass is 75kg and the gravitational field strength on Earth is 10N/kg.
W = 75 kg x 10 N/kg
W = 750 N
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By how much do you stretch your 0.600-cm diameter nylon rope when you hang 23.0m
below a rock outcropping. Assume your mass is 75.0 kg and your nylon rope has Young’s modulus of 5.00 × 10^9 N/m^2.
Answer:
To solve for the amount of stretch in the nylon rope, we can use the equation for the elongation (stretch) of a rope under tension:
ΔL = FL / AE
where ΔL is the change in length of the rope, F is the force on the rope, L is the original length of the rope, A is the cross-sectional area of the rope, and E is the Young's modulus of the rope.
First, we need to find the force on the rope. This is equal to the weight of you and the rope:
F = mg = (75.0 kg + mass of rope)g
Next, we need to find the cross-sectional area of the rope. The diameter of the rope is given as 0.600 cm, so the radius is 0.300 cm = 0.00300 m. Therefore, the cross-sectional area is:
A = πr^2 = 2.83 × 10^-6 m^2
Now we can plug in the given values and solve for ΔL:
ΔL = FL / AE = [(75.0 kg + mass of rope)g](23.0 m) / [(2.83 × 10^-6 m^2)(5.00 × 10^9 N/m^2)]
We can simplify this expression by using the fact that the mass of the rope is much smaller than your mass, so we can assume that the force on the rope is equal to your weight:
ΔL = (mg)(23.0 m) / (A E) = (75.0 kg)(9.81 m/s^2)(23.0 m) / [(2.83 × 10^-6 m^2)(5.00 × 10^9 N/m^2)]
Solving for ΔL, we get:
ΔL = 0.068 cm
Therefore, the nylon rope stretches by approximately 0.068 cm when you hang 23.0 m below a rock outcropping.
During the earliest stages of the universe, the only things that existed were
During the earliest stages of the universe, the only things that existed were subatomic particles such as protons, neutrons, and electrons. These particles came together to form the first atoms, which were primarily hydrogen and helium.
The universe was also filled with a hot, dense plasma of particles and radiation, known as the cosmic microwave background radiation.
Hence, as the universe expanded and cooled, these atoms and radiation would play a key role in the formation of galaxies, stars, and the larger structures we see today.
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Suppose the metal sample in Figure 10.2 is immersed in water, is cubical with side 3.0 cm, and has a mass of 54 g.
The force of buoyancy acting on the sample is 2.65 N.
If the sample's supporting string is cut, it will float to the surface of the water.
How to calculate bouyant force?a. The buoyant force acting on the sample can be calculated using the formula Fb = ρVg, where ρ is the density of water, V is the volume of the submerged part of the sample, and g is the acceleration due to gravity. Since the sample is immersed in water, its volume is equal to the volume of water it displaces, which can be calculated using the formula V = l × w × h, where l, w, and h are the dimensions of the submerged part of the sample. Since the sample is cubical, all sides have the same length of 3.0 cm. Thus, the submerged part of the sample has a volume of V = 3.0 cm × 3.0 cm × 3.0 cm = 27.0 cm³ = 0.027 m³. The density of water is ρ = 1000 kg/m³, and the acceleration due to gravity is g = 9.81 m/s². Therefore, the buoyant force acting on the sample is:
Fb = ρVg = 1000 kg/m³ × 0.027 m³ × 9.81 m/s² ≈ 2.65 N
b. The sample will float to the surface of the water if its supporting string is cut. This is because the buoyant force acting on the sample is greater than its weight, which means that there is a net upward force on the sample. This net force causes the sample to accelerate upward, and it will continue to accelerate until it reaches the surface of the water. At the surface, the upward force is balanced by the weight of the sample, and it will float on the water.
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Complete question:
2. Suppose the metal sample in Figure 10.2 is immersed in water, is cubical with side 3.0 cm, and has a mass of 54 g. a. Calculate the buoyant force acting on the sample. b. Describe the behavior of the sample if its supporting string is cut. Explain how you arrive at your answer. balance arm water metal sample water overflow can beaker Figure 10.2 MEASUREMENTS
Two batteries supply current to the circuit in the figure. The figure shows the potential difference across two of the resistors and the value of the third resistor.(Figure 1). What current is supplied by the batteries?
The current that is supplied by the batteries is 0.01333 Amp.
How to explain the informationAn electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor. It is measured as the net rate of flow of electric charge through a surface or into a control volume.
Eeq = E1 + E2 ( batteries in series )
=> Eeq = 3+4.5 = 7.5 V
Veq = V1 + V2 + V3 ( resistances in series )
=> V3 = 7.5 - 2 - 3.5 = 2 V
=> current = V/R = 2/150 = 0.01333 Amp
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A 60.6 kg weight-watcher wishes to climb a mountain to work off the equivalent of a large piece of chocolate cake rated at 720 (food) Calories. How high must the person climb? The acceleration due to gravity is 9.8 m/s 2 and 1 food Calorie is 103 calories. Answer in units of km.
The height the weight-watcher needs to climb to work off the equivalent of the chocolate cake is 0.5225 km.
What is the height of the person?The height the weight-watcher needs to climb to work off the equivalent of the chocolate cake is calculated as follows;
P.E = mgh
Where;
m is the mass of the objectg is the acceleration due to gravityh is the height.720 Calories x 103 calories/1 food Calorie = 74,160 calories
74,160 calories = 310285.44 J
h = P.E/mg
h = (310285.44) / (60.6 x 9.8)
h = 522.5 m
h = 0.5225 km
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If enough experimental data supports a hypothesis, then it:
A. Is accepted as true until proven false.
B. Becomes an Observational Law
C. Is proven 100% true.
D. Is falsified.
If enough experimental data supports a hypothesis, it is considered a well-supported scientific theory, but it is not considered to be 100% true or proven. Scientific theories are always open to further investigation and revision based on new evidence. Therefore, option C ("Is proven 100% true") is incorrect.
Option A ("Is accepted as true until proven false") is also incorrect because scientists do not accept a hypothesis as true until it has been rigorously tested and supported by a large body of evidence. Even then, scientists recognize that any scientific theory is subject to revision or falsification if new data or evidence emerges that contradicts it.
Option B ("Becomes an Observational Law") is also incorrect because scientific laws are typically descriptive, rather than explanatory. They describe what happens in a given set of circumstances, but they do not explain why it happens. Hypotheses and theories, on the other hand, attempt to explain why certain phenomena occur, and they are supported by experimental evidence.
Therefore, none of the options are completely accurate, but the most appropriate answer is that the hypothesis becomes a well-supported scientific theory.
of the following stars, which one has a mass less than that of our sun? The choices are Sirius B, Altair, Arcturus, Spica, and Rigel.
Answer:
Sirius BExplanation:
Here we go !!!!
Sirius B: A white dwarf star and companion to Sirius A, the brightest star in the sky. It has a mass of only about 0.98 times that of the sun, making it less massive than our sun.Arcturus: A red giant star located in the constellation Boötes, with a mass of about 1.1 times that of the sun.Spica: A binary star system composed of two hot, blue stars located in the constellation Virgo. The primary star has a mass of about 11 times that of the sun.Rigel: A blue supergiant star located in the constellation Orion, with a mass of about 17 times that of the sun. It is one of the brightest stars in the sky and is a part of Orion's Belt.Altair: A main-sequence star of spectral type A7V, located in the constellation Aquila. It has a mass of about 1.8 times that of the sun, which means it is more massive than the sun.I hope this information helps!
The center of gravity of a 5.00 kg irregular object is shown in (Figure 1). You need to move the center of gravity 1.60 cm to the left by gluing on a 1.50 kg mass, which will then be considered as part of the object.
Where should the center of gravity of this additional mass be located?
The moment of the weight of the 1.50 kg mass about the CG of the 5.00 kg object will be considered as part of the object.
The center of gravity of the additional 1.50 kg mass should be located 1.60 cm to the right of the current CG of the 5.00 kg object.
What is the moment ?
We can use the principle of moments to determine the location of the center of gravity of the additional 1.50 kg mass that needs to be glued on.
The principle of moments states that the sum of the moments of all the forces acting on an object is equal to the moment of the resultant force about any point. In this case, we can take moments about the current center of gravity (CG) of the 5.00 kg object to determine the location of the center of gravity of the additional 1.50 kg mass.
The moment of a force about a point is given by the product of the magnitude of the force and the perpendicular distance from the point to the line of action of the force. Since the gravitational force acting on the masses is the only force acting on the system, we can take moments of the weight of each mass about the CG of the 5.00 kg object.
The moment of the weight of the 5.00 kg object about its CG is zero, since the CG is the point about which the object is in rotational equilibrium. Therefore, we only need to consider the moment of the weight of the 1.50 kg mass about the CG of the 5.00 kg object.
The distance between the CG of the 5.00 kg object and the desired new CG is 1.60 cm to the left, so we need to find the distance between the current CG and the new CG. Let x be the distance from the current CG to the new CG.
The moment of the weight of the 1.50 kg mass about the current CG is:
(1.50 kg)(9.81 m/s²)(x) = 14.72x Nm
The moment of the weight of the 1.50 kg mass about the new CG is:
(1.50 kg)(9.81 m/s²)(1.60 cm + x) = 14.72x + 23.54 Ncm
Setting these two moments equal, we have:
14.72x Nm = 14.72x + 23.54 Ncm
Converting the units, we get:
0.1472x m = 0.2354 m
x = 1.60 cm (to two significant figures)
Therefore, the center of gravity of the additional 1.50 kg mass should be located 1.60 cm to the right of the current CG of the 5.00 kg object.
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A blue train of mass 50 kg moves at 4 m/s toward a green train of 30 kg initially at rest. The trains collide. After the collision the green train moves with a speed of 3 m/s. What is the final momentum of the green train?
A. 200 kgm/s
B. 90 kgm/s
C. 20 kgm/s
D. 110 kgm/s
Determine the percentage of kinetic energy lost by a small ball when it makes an elastic head-on collision with stationary bigger ball. The mass of the bigger ball is 12 times bigger than the mass of the small ball.
The small ball loses 2.37% of its kinetic energy during the elastic head-on collision with the stationary bigger ball.
[tex]\frac{Kf}{Ki} =[/tex] [tex]1- [\frac{1}{2}][ \frac{M}{m}] [\frac{2mv^{2} }{[M+m]^{2} }[/tex]
M = 12m
Kf/Ki = 1 - (1/2)(12m/m)[(2mv^2)/(13m)^2]
Kf/Ki = 165/169
(1 - Kf/Ki) x 100% = (1 - 165/169) x 100% = 2.37%
So, the small ball loses 2.37% of its kinetic energy during the elastic head-on collision with the stationary bigger ball.
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pls help in astronomy didn’t know what subject to put it under
The subject depicted in the attached image is Astronomy and Astrophysics.
Definitely younger than the SunAO main sequence starB-type starsF-type stars (some)Possibly younger than the SunF1 main sequence starG2 main sequence starMO main sequence starDefinitely older than the SunM-type stars (some)M1, 1 Msun red giantM1, 18 Msun red supergiantWhat is Astronomy?Astronomy is the scientific study of celestial objects such as stars, planets, galaxies, and other phenomena that exist outside of Earth's atmosphere.
Astronomers use a variety of methods to observe and study these objects, including telescopes, spacecraft, and computer simulations.
Astronomy is a broad field that includes many different sub-disciplines, such as astrophysics, planetary science, and cosmology.
Astronomers study the physical properties and behavior of celestial objects, such as their composition, temperature, motion, and evolution.
They also seek to understand the structure and history of the universe as a whole.
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A hydraulic system contains one small piston that has a diameter of 1 (one) inch, and a large piston that has an area of 12.56 in. What is the system pressure if the large piston produces a force of 37,775 pounds?
The pressure of the system, that the large piston produces a force of 37775 pounds is 3007.56 pound / in²
How do i determine the pressure of the system?The following data were obtained from the question:
Area of large piston = 12.56 in² Force of large piston = 37775 poundsPressure of system =?Pressure is defined as force per unit area as shown by the following formula
Pressure = Force / Area
Inputting the value of the force and area, we have
Pressure of system = 37775 / 12.56
Pressure of system = 3007.56 pound / in²
Thus, from the calculation made above, we can conclude that the pressure of the system is 3007.56 pound / in²
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a) A 0.250-kg toy car is propelled by a compressed spring. The car follows a track that rises above the starting point. The spring is compressed 6.00 cm and has a force constant of 450.0 N/m. Find the speed of the car when it reaches 0.110 m above the starting point.
b) Estimate the power expended by a child raising a 0.25-kg toy 0.75 m in 2.0 s. Convert your answer to horsepower (1 ℎp = 746w)
Answer:
a) To solve for the speed of the car, we can use the conservation of energy equation:
(1/2)mv^2 = mgh - (1/2)kx^2
where m is the mass of the car, v is the speed of the car, g is the acceleration due to gravity, h is the height of the car above the starting point, k is the spring constant, and x is the distance the spring is compressed.
Plugging in the given values, we get:
(1/2)(0.250 kg)v^2 = (0.250 kg)(9.81 m/s^2)(0.110 m) - (1/2)(450.0 N/m)(0.0600 m)^2
Solving for v, we get:
v = 1.23 m/s
Therefore, the speed of the car when it reaches 0.110 m above the starting point is 1.23 m/s.
b) To estimate the power expended by a child raising a 0.25-kg toy 0.75 m in 2.0 s, we can use the work-energy theorem:
W = ΔE = mgh
where W is the work done, m is the mass of the toy, g is the acceleration due to gravity, h is the height the toy is raised, and ΔE is the change in energy.
Plugging in the given values, we get:
W = (0.25 kg)(9.81 m/s^2)(0.75 m) = 1.84 J
Next, we can use the definition of power:
P = W/t
where P is power, W is work, and t is time.
Plugging in the given values, we get:
P = 1.84 J / 2.0 s = 0.92 W
Finally, we can convert watts to horsepower:
1 hp = 746 W
0.92 W = 0.92 / 746 hp = 0.0012 hp
Therefore, the power expended by a child raising a 0.25-kg toy 0.75 m in 2.0 s is approximately 0.0012 hp.
Which of the following can be formed by a convergent boundary?
A. Volcanic mountains B. All of these
C. Volcanic islands
D. Deep ocean trenches
Answer:
Volcanic mountains can be formed by a convergent boundary.
* Find out the force between the wires 150 cm long and I'm apart) used to connect the battery of a car to the motor? [Note: The Current in the wires in 200 ampere)
Answer:
The force between the wires used to connect the battery of a car to the motor can be calculated using the equation F = I*L, where F is the force, I is the current, and L is the length of the wires. In this case, the force would be calculated as F = 200A * 0.15m = 30N.
Find the ratio of the coulomb electric force Fe to the gravitational force Fg between two electrons in vacuum.
5. How mach work is repuried to increase the speed of 4kg object from 20m/s to 40m/s
math
Determine if the function below is continuous.
graph of a piecewise function, with 2 pieces. The first piece is a line that starts at negative infinity and goes through point (-4,5) and ends with an open dot at (1,0). The second piece is a line that starts with a closed dot at (1,-1), goes through point (2,1) and continues to infinity.
A. not continuous at x = 1
B. not continuous at x = 0
C. not continuous at x = -1
D. continuous
The function is not continuous at x=1, so the answer is A.
At x=1, the two pieces of the function meet. The first piece ends with an open dot at (1,0), meaning that the function is not defined at $x=1$. The second piece starts with a closed dot at (1,-1), meaning that the function is defined at x=1 and takes the value -1 there.
Since the function is not defined at x=1, it cannot be continuous at that point. Therefore, the function is not continuous overall.
Sources of error could have come from friction that may result in energy losses during the collision, The gliders may not be perfectly elastic, which means that some energy may be lost during the collision
are these errors random or systematic errors?
Sources of error could have come from small amounts of friction, and glider 2 could have not been totally at rest. These errors are systematic errors.
Are these errors random or systematic?
Answer:
The first set of errors, which include friction and imperfect elasticity, are systematic errors because they arise from consistent factors that affect the measurements in a predictable way. These errors will be present in every trial of the experiment and will cause a consistent deviation from the true value.
The second set of errors, which include small amounts of friction and the initial velocity of glider 2, are also systematic errors because they arise from consistent factors that affect the measurements in a predictable way. These errors will also be present in every trial of the experiment and will cause a consistent deviation from the true value.
Explanation:
A piece of wood of density 400kg/m³ lump of aluminium is tied to a lump of aluminium of mass 0.01kg and density of 270kg/m³ . the arrangement has a mean density of 100kg/m³and just float in water. Determine the just float in volume of the piece of wood
The just float in volume of the piece of wood is 0.008 cubic meters.
Just float calculationLet's denote the volume of the piece of wood as V_wood, and the volume of the lump of aluminum as V_aluminum.
We can start by using the information given to write two equations:
Equation 1: (density of wood) * V_wood + (density of aluminum) * V_aluminum = (mean density) * (V_wood + V_aluminum)
Substituting the given values, we get:
400 * V_wood + 270 * V_aluminum = 100 * (V_wood + V_aluminum)
Simplifying and rearranging, we get:
300 * V_wood = 70 * V_aluminum
Equation 2: The total mass of the arrangement is the sum of the masses of the wood and aluminum:
(density of wood) * V_wood + (density of aluminum) * V_aluminum = (total mass)
Substituting the given values, we get:
400 * V_wood + 270 * V_aluminum = 0.01
Now we can use Equation 1 to eliminate V_aluminum from Equation 2:
300 * V_wood = 70 * V_aluminum
V_aluminum = (300/70) * V_wood
Substituting this into Equation 2, we get:
400 * V_wood + 270 * [(300/70) * V_wood] = 0.01
Simplifying and solving for V_wood, we get:
V_wood = 0.008 m^3
Therefore, the just float in volume of the piece of wood is 0.008 cubic meters.
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Can u please help with first two tasks
Workbook:complete physics for Cambridge secondary first
Electromagnetic induction is a phenomenon in physics that occurs when a varying magnetic field produces an electric current or voltage in a conductor. It was first discovered by Michael Faraday in the early 19th century and is now a fundamental principle of modern electrical technology.
When a conductor, such as a wire, is moved through a magnetic field, or when the magnetic field itself changes, a voltage is induced across the ends of the conductor. The magnitude of the induced voltage is proportional to the rate of change of the magnetic field and the number of turns of wire in the coil.
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Explain why sound wave travel faster in liquids than in gases
Answer:
Sound travels faster in liquids than in gases because molecules are more tightly packed. In fresh water, sound waves travel at 1,482 meters per second (about 3,315 mph). That's well over 4 times faster than in air!
Explanation:
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The surfaces of a lipid bi-layer forming the membrane around a cell with a radius of 1.5 μm has a residual charge qr = 9.1×10-15 C on outside of the bi-layer, and the same amount of negative charge on the inside. What is the force in pN (×10-12 N) on a singly-charged positive ion (q =1.6 ×10-19 C) located on the outer surface of this membrane? Hint: Use F = q E = q (σ/e) with σ = qr/A = qr/ (4π r2) and εo = 8.85 x 10-12 F·m-1.
I mean if it will be anything the asnwer would be 23^90'200
URGENT 50 POINTS
Show all Steps of your work.
7. Mohammed whose mass is 50kg swings back and forth on a long vine makes an angle 45 from the vertical resting position. His friend Abdella notices in amazement that he makes 30 complete swings in 2.0 minuies.
(a) What is the frequency (in hertz) of Tarzan's swing?
(b) What is the period of oscillation?
(c) How long is the vine he is using?
(d) Calculate the Restoring force on Mohammed?
The frequency of oscillation of the swing is 0.25 Hz.
The time period of oscillation of the swing is 4 s.
The length of the vine of the swing is 3.97 m.
The restoring force acting on Mohammed is 692.9 N.
Mass of Mohammed, m = 50 kg
Angle made by the vine with the vertical, θ = 45°
Number of complete swings made by Mohammed, n = 30
Time taken for this swing, t = 2 minutes = 120 seconds
a) The frequency of the swing is defined as the number of complete oscillations in one second.
So, the frequency of oscillation of the swing is,
f = n/t
f = 30/120
f = 0.25 Hz
b) The time period of oscillation of the swing is,
T = 1/f
T = 1/0.25
T = 4 s
c) The expression for the time period is given by,
T = 2π√(l/g)
T² = 4π² x (l/g)
l/g = T²/4π²
Therefore, the length of the vine of the swing is,
l = T²g/4π²
l = 4² x 9.8/4 x (3.14)²
l = 3.97 m
d) The restoring force acting on Mohammed,
F = mg sinθ
F = 50 x 9.8 x sin 45°
F = 490 x 1/√2 = 490/1.414
F = 692.9 N
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A baseball player wants to hit a home run over the wall of a
stadium. The player swings the baseball bat so that it hits the
ball when it is at a height of 0.996 m above the ground. The
ball flies off at an angle of 30° above the horizontal and at a
speed of 36.2 m/s. What is the tallest wall that the player can
clear (i.e., get the ball over) if the wall is 99.1 m away
horizontally?
A wall less than or equal to 14.7 m in height is the highest the player can clear.
How to calculate height?Use the kinematic equations of motion to solve the problem. First, we can find the time it takes for the ball to travel 99.1 m horizontally:
d = vt
t = d / v
t = 99.1 m / 36.2 m/s
t ≈ 2.74 s
Now, use the vertical motion equation to find the maximum height the ball reaches:
y = yo + vot + (1/2)at²
where:
yo = 0.996 m (initial height)
vo = v sinθ = 36.2 m/s x sin(30°) ≈ 18.1 m/s (initial vertical velocity)
a = -9.81 m/s² (acceleration due to gravity, pointing downward)
t = 2.74 s (time of flight)
y = 0.996 m + 18.1 m/s x 2.74 s + (1/2) x (-9.81 m/s²) x (2.74 s)²
y ≈ 14.7 m
Therefore, the tallest wall the player can clear is a wall with a height less than or equal to 14.7 m.
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