The density of the mercury in lbs/l is 29.96 lbs/l.
Density: The ratio of the mass of the substance to its total volume is defined as the density of that substance.
The formula of the density is,
d=m/V
where d is the density, m is the mass and V is the volume.
Conversion of the unit from g/ml to lbs/l:
It is given that the density of the given substance is 13.6 g/ml.
1 lb = 0.454 Kg
1 kg = 1000 g
So 1 lb = 0.454 *1000 g
1 lb = 454 g
That means 1 g = 1/454 lb
Then 1 ml = 0.001 l.
Using these values, the density of the given substance in lbs/l is calculated.
13.6 g/ml = (13.6 g/ml)*(1 lb/ 454 g)÷(0.001 l/ml)
13.6 g/ml = 29.96 lbs/l
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Suppose the carts collided on a surface that had a slight incline to it. Would you expect the momentum to be conserved
Momentum is conserved when carts are collided on a slanting plane.
To find the answer, we need to know about the conversation of momentum.
What's the conversation of momentum?Conservation of linear momentum says the total momentum before the collision and after the collision remains the same. Mathematically, m1u1+m2u2 = m1v1+m2v2How is the momentum conserved when collision occurs on a slanting plane?On a slanting plane, the velocity has two components, horizontal component horizontal component Vertical componentSo, its momentum has also similar two components. The momentum is conserved along horizontal direction and vertical direction separately.Thus, we can conclude that the momentum is conserved when carts are collided on a slanting plane.
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A wire is joined to points X and Y in the circuit diagram shown. A diagram of a circuit with a power source on the left. Directly above the power source is a dot labeled X and then a circle with an X in it. The circuit then splits with one path straight back to the power source and the other path has 3 circles with X in them labeled 2, 3, and 4 respectively. There is a point labeled Y between circles 2 and 3.A diagram of a circuit with a power source on the left. Directly above the power source is a dot labeled X and then a circle with an X in it. After the x are 4 different circles on the circuit with Xs in them labeled 1, 2, 3, and 4 respectively. There is a point labeled Y in between circles 2 and 3. There is a branch of the circuit from X to Y. How does the circuit change when the wire is added?
A closed circuit occurs and makes all bulbs turn off.
An open circuit occurs and makes all bulbs turn off.
A short circuit occurs and makes bulbs 3 and 4 turn off but keeps bulbs 1 and 2 lit.
A short circuit occurs and makes bulbs 1 and 2 turn off but keeps bulbs 3 and 4 lit.
The circuit change when the wire is added will see a short circuit occur and makes bulbs 1 and 2 turn off but keeps bulbs 3 and 4 lit. Option D. This is further explained below.
How does the circuit change when the wire is added?Generally, Electronic circuits consist of a series of interconnected parts that form a closed loop through which electricity may flow.
In conclusion, If two wires are linked together, a short circuit will develop, cutting power to bulbs 1 and 2. But there is no impact on bulbs 3 and 4. There is no problem with bulbs 3 and 4.
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A uniform stick has a mass of 4.42 kg and a length of 1.23 m. It is initially lying flat at rest on a frictionless horizontal surface and is struck perpendicularly by a puck imparting a horizontal impulsive force of impulse 12.8 N s at a distance of 46.4 cm from the center. Determine the subsequent motion of the stick.
10.7 rad/s is the final angular velocity of the stick.
Given:
Mass of the stick = 4.42 kg
Length of the stick = 1.23m
Force of impulse (I) = 12.8 N s
The linear velocity of the stick, [tex]v=\frac{I}{m}[/tex]
[tex]v=\frac{12.8 N.s (\frac{1 kg m/s^2}{1 N}) }{4.42 kg}[/tex]
[tex]v[/tex] [tex]= 2.89 m/s[/tex]
Therefore, the final linear velocity of the stick is 2.89 m/s
∴[tex]w=\frac{12 Ir}{ml^{2} }[/tex]
[tex]w=\frac{12 ( 12.8 N.s ) ( 46.4 cm)}{(4.42 kg) (1.23 m)^2}[/tex]
[tex]w= \frac{12 (12.8 N.s) (46.4 cm) (\frac{10^-^2 m}{1 cm} )}{(4.42 kg) (1.23m)2}[/tex]
[tex]w=10.7 rad/s[/tex]
Therefore, 10.7 rad/s is the final angular velocity of the stick.
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With a mass of 111.1 kg, is moving with a velocity of 3.000 m/s to the right. Mass is initially at rest and has a mass of 222.2 kg. After the elastic collision, bounces off with a velocity of 1.000 m/s to the left. Calculate the velocity of after the collision.
Let us assume the positive \hat{x} direction is towards the right.
During any collision, the total momentum is conserved.
That means, the total momentum before the collision is the same as the total momentum after the collision.
p before = P after
Given,
Mass 1, m_1 = 111.1kg
Mass 2, m_2 = 222.2kg
Initial Velocity of Mass 1, 1,1 3.00.î m/s
Initial Velocity of Mass 2, \overrightarrow v_{2,i} =0
Velocity of Mass 1 after collision, \overrightarrow v_{1,f} =-1.000\hat{x}\hspace{0.1cm}m/s
The total Momentum of the system before the collision is,
\overrightarrow p_{before} =m_1\overrightarrow v_{1,i} + m_2\overrightarrow v_{2,i}
=> \overrightarrow p_{before} =(111.1)(3.00\hat{x}) + 0
=> \overrightarrow p_{before} =333.3\hspace{0.1cm}\hat{x}\hspace{0.1cm}kgm/s
The total Momentum of the system after the collision is,
\overrightarrow p_{after} =m_1\overrightarrow v_{1,f} + m_2\overrightarrow v_{2,f}
=> \overrightarrow p_{after} =(111.1)(-1.000\hat{x}) + (222.2)\overrightarrow v_{2,f}
=> \overrightarrow p_{after} =(-111.1\hat{x}) + (222.2)\overrightarrow v_{2,f}
We know the total momentum before the collision is the same as the total momentum after the collision.
p before = P after
=>333.3\hat{x} =(-111.1\hat{x}) + (222.2)\overrightarrow v_{2,f}
=>(-111.1\hat{x}) + (222.2)\overrightarrow v_{2,f} = 333.3\hat{x}
=> (222.2)\overrightarrow v_{2,f} = 333.3\hat{x} + 111.1\hat{x}
=> (222.2)\overrightarrow v_{2,f} = 444.4\hat{x}
=> \overrightarrow v_{2,f} = \frac{444.4}{222.2}\hspace{0.1cm}\hat{x}\hspace{0.1cm}m/s
=> \overrightarrow v_{2,f} = 2.00\hspace{0.1cm}\hat{x}\hspace{0.1cm}m/s
Therefore the velocity of m_2 after collision is v_{2,f} = 2.00\hspace{0.1cm}m/s going to the right
Please check the attached file for a brief answer.
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Verification of laws of reflection of sound (Record Work)
The Angle of Incidence is equal to the Angle of Reflection.
What is the Verification of laws of reflection of sound?Take a wooden drawing board and fix a white sheet of paper on it. In the middle of paper draw a straight line and Mark a point B on it. Draw a perpendicular . Place a mirror on line such that the flat side of the mirror is along the line. Hold the mirror in the mirror holder.
Fix two steel pins P and Q on the straight line AB at least 10 cm apart. Look for the images of the pins P and Q and fix two pins P a such that P', Q', and images of P and Q are all in the same straight line. Remove the pins and draw small circles around the pinpricks.
Remove the mirror and Join P'Q' which produce the straight line to meet at B. Measure ∠ABN = i and ∠CBN = r. It is found that ∠i=∠r. This proves that the Angle of Incidence is equal to the Angle of Reflection.
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Assume the first dot for animal A represents t = 0. At which time does animal B overtake animal A?
The time in which the animal B overtake animal A according to the motion map is at 2 seconds.
What is a Motion map?This is referred to as one-dimensional plots built off of a position line and depicts velocity and acceleration.
The time in which Animal B is ahead of animal A is at 2 seconds which therefore makes it the most appropriate choice.
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a car travelling at 5m/s is brought to rest in a distance of 20m.Find it's decelaration and time taken.
Answer:
time taken= 8s
deceleration = -0.625ms^-2
Explanation:
s = 1/2(u+v)t (linear motion formula)
20 = 1/2 ( 5+0 ) t
t=time taken, find t
t= 8s
a= (v-u) / t
a= (0-5) / 8s
= -0.625 ms^-2
What is weight??????????????????
Answer:
Weight is a body's relative mass, or the quantity of matter contained by it.
Explanation:
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1. A large ball was let go on a hill and started rolling down with a constant acceleration of 4.2 m/s². What was the velocity of the ball after 12 s?
Answer:
The velocity after 12s is 50.4m/s.
Explanation:
In acceleration formula make velocity the subject.
acceleration(a) = velocity(v)÷time(t)
velocity (v) = acceleration(a)×time(t)V = 4.2m/s²×12s
V = 50.4m/s
Therefore the velocity after 12s is 50.4m/s.
which process receives geo thermal energy naturally
A type of electromagnetic radiation that has a higher frequency than visible light is Question 15 options: ultraviolet infrared radiowaves microwaves two of the above
Ultraviolet rays are Electromagnetic waves that have higher frequencies than visible light, so they carry more energy.
What are Ultraviolet rays?Johann Wilhelm Ritter, a German physicist, first noticed that paper soaked in silver chloride browned more quickly when exposed to Ultraviolet rays than when it was exposed to violet light itself in 1801.For the most part, humans cannot see ultraviolet rays. The cornea filters out shorter wavelengths, while the lens of the human eye primarily inhibits radiation in the 300–400 nm range. Additionally, UV color receptor adaptations are not present in humans. The retina's photoreceptors are sensitive to near-UV, though, and those without lenses (a condition known as aphakia) see near-UV as whitish-blue or whitish-violet.The body produces vitamin D, which is necessary for life when exposed to UV light, more especially UVB. UV rays are necessary for humans to maintain proper vitamin levels.To learn more about Electromagnetic waves refer to:
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g Three projectiles are launched from the same level on flat ground. Each reaches the same maximum height before returning to the ground, but projectile 3 (in blue) has the largest range, followed by projectile 2 (in green). Projectile 1 has the smallest range (in red). Rank the three from largest to smallest according to:
A stone of mass 150g is rotated in a horizontal circle at 10m/s which is attached to the end of a 1m long. what will be the acceleration of the stone and it's centripetal force?
force is mass multiply by acceleration so it will be 150 multiply by 10 is 1500N
Answer:
Acceleration: [tex]100\; {\rm m\cdot s^{-2}}[/tex] assuming that the radius of the rotation is [tex]1\; {\rm m}[/tex].
Centripetal force: [tex]15\; {\rm N}[/tex].
Explanation:
In a circular motion, if the tangential velocity is [tex]v[/tex] and the radius of the motion is [tex]r[/tex], the centripetal acceleration of the motion would be [tex]a = (v^{2} / r)[/tex].
In this question, it is implied that for this circular motion, [tex]v = 10\; {\rm m\cdot s^{-1}}[/tex] while [tex]r = 1\; {\rm m}[/tex]. Thus, the (centripetal) acceleration would be:
[tex]\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(10\; {\rm m\cdot s^{-2}})^{2}}{1\; {\rm m}} \\ &= 100\; {\rm m \cdot s^{-2}}\end{aligned}[/tex].
Note that the unit of mass in this question is gram, whereas the standard unit for mass should be [tex]{\rm kg}[/tex] (so as to leverage the fact that [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex].) Apply unit conversion: [tex]m = 150\; {\rm g} = 0.150\; {\rm kg}[/tex].
Using that fact that [tex](\text{net force}) = (\text{mass}) \, (\text{acceleration})[/tex]:
[tex]\begin{aligned} (\text{net force}) &= (\text{mass}) \, (\text{acceleration}) \\ &= 0.150\; {\rm kg} \times 100\; {\rm m\cdot s^{-2}} \\ &= 15\; {\rm kg \cdot m \cdot s^{-2}} \\ &= 15\; {\rm N}\end{aligned}[/tex].
A substance that has 1500 liters and which has a density of 250 kg / m3 determine the mass of said substance
【Answer】Therefore, the mass in said substance is 375 kilograms.
[tex]\red{ {\hspace{50 pt}\above 1.2pt}\boldsymbol{\mathsf{Procedure}}{\hspace{50pt}\above 1.2pt}}[/tex]
This is an exercise on fluids and their fundamental characteristics.
We start to solve, obtaining the data:
DATA:v = 1500 lt = 1.5 m³d = 250kg/m³m = ?Conversion from liters to m³
[tex]\boldsymbol{1500\not{l}*\dfrac{1 \ m^{3} }{1000\not{l} }=1.5 \ m^{3} }[/tex]
To calculate mass: multiply density by volume.
[tex]\boldsymbol{m=d*v \ \ \to \ \ \ Formula}[/tex]
We clear our data in the formula:
[tex]\boldsymbol{m=250\dfrac{kg}{\not{m^{3}}}*1.5\not{m^{3}} }[/tex]
[tex]\boldsymbol{m=375 \ kg}[/tex]
{ Pisces04 }A box weighing 660 N is sliding across a cement floor and the coefficient of sliding
friction between the box and the floor is 0.15. If the force pushing the box is 500 N, what
is the acceleration of the box?
O 6 m/s²
O 2 m/s²
O 10 m/s²
pts
O 9 m/s²
The acceleration of the box is 6 m/s²; option A.
What is the acceleration of the box?The acceleration of the box is determined as follows:
Frictional force = 660 * 0.15 = 99 N
Net force = 500 - 99 = 401 N
Force = mass * accelerationMass of the box = 660/9.8 = 67.3 kg
Acceleration = net force/mass
Acceleration = 401/67.3
Acceleration = 6 m/s²
In conclusion, the acceleration of the box is determined from the net force and the mass of the box.
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A light bulb receives 100 J of energy electrically. It transfers 30 J of energy to the surroundings by heat. What is the efficiency of the light bulb?
Answer:
The efficiency of the light bulb is 30%.
Explanation:
Efficiency = output/input × 100%
Data
Energy input = 100J
Energy output = 30J
Efficiency = ?
Efficiency = Energy output/Energy input × 100%
Efficiency = 30J/100J × 100%
Efficiency = 0.3 × 100%
Efficiency = 30%
Therefore the efficiency of the light bulb is 30%.
which two forms of energy do burning substances produce
Answer:thermal (heat) energy and radiant (light) energy.
Explanation:
Answer:
Answer: We use the chemical energy in fuels by burning them and transforming them into other types of energy: thermal energy, as when we burn fuel for heat; and kinetic energy, as when we burn gasoline to power our car's motion
Explanation:
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If oxygen, which has a electronegativity of 3.5, bonds with hydrogen, which has an electronegativity of 2.1, the bond between the two atoms will be classified as a polar____ bond
Given the electronegativity of O is 3.5 and H is 2.1, the bond between the two atoms will be classified as a polar covalent bond.
How are bonds classified?According to the difference in electronegativity (ΔEN).
If ΔEN ≤ 0.4, the bond is nonpolar covalent.If 0.4 < ΔEN ≤ 1.9, the bond is polar covalent.If ΔEN > 1.9, the bond is ionic.Oxygen has an electronegativity of 3.5 and hydrogen of 2.1. The difference in electronegativity is:
ΔEN = EN(O) - EN(H) = 3.5 - 2.1 = 1.4
Given the electronegativity of O is 3.5 and H is 2.1, the bond between the two atoms will be classified as a polar covalent bond.
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If the distance between two charged particles is increased 2.69 times, the ratio of new to old electric force is _______ times the old value. (Round your answer to two decimal places).
Answer:
F = K Q1 Q2 / R^2 force between 2 charged partices
F2 / F1 = (R1 / R2)^2 = (1 / 2.69)^2 = .139
F2 = .139 F1
If the current through a resistor is tripled, how does the power dissipated by
the resistor change?
If the current that is circuit is tripled, the power is increased by a factor of 9. Option D
What is power?The term power refers to the rate of doing work. Now if we have to obtain the work done by the use of the formula; P = I^2R
Now if the current that is circuit is tripled, the power is now obtained by;
P = (3I)^2R = 9I^2R hence the power is increased by a factor of 9.
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All objects fall at the same rate, what is the acceleration of the force of gravity?
Answer:
9.8 m/s
Explanation:
If these objects are in free fall, meaning that their downward movement is only due to the force of gravity, then acceleration due to gravity is about 9.8 metres per second per second.
In an antique automobile, a 6-V battery supplies a total of 48 W to two identical headlights in parallel. The resistance of each bulb is:
Answer:
The resistance of the each bulb is 1.75 ohm.
Explanation:
The Power P supplied by the battery is given by the formula,
P=VI
where V is the voltage and I is the current.
Then the total current through given parallel circuit is I=P/V
Given P=48 W and V=6 V the value of I is
I=48 W/6 V
I= 8 A
Since both bulbs have same resistance and are connected in parallel so the current through each circuit will be half of the total current. Hence the current through each circuit I₀ is 4 A.
The Resistance R is calculated using formula R=V/I₀.
So the resistance is
R=6 V/4 A
R=1.75 ohm
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Answer:
The resistance of the bulb is 1.5 ohm.
Explanation:
The power P delivered to the whole circuit is given by the formula,
P=VI
Given P= 48 W and V= 6V, the current in the circuit is
I=P/V
I=48/6= 8 A.
Since the bulbs are identical and are in parallel, therefore the current through each parallel circuit is half of the total current. Therefore current I1 through each bulb is 4 A.
The resistance is given by the formula,
R=V/I1
Hence R=6/4 = 1.5 ohm
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. A microwave oven operates at 2.4GHz with an intensity inside the oven of 2500 W/m^2. What are the amplitudes of the oscillating electric and magnetic fields
Answer: A microwave oven operates at 2.4GHz with an intensity inside the oven of 2500 W/m^2. Then, the amplitudes of the oscillating electric and magnetic fields are, [tex]B_0=4.57*10^{-6} T and E_0=1.37*10^{3} V/m[/tex]
Explanation: To find the correct answer, we have to know more about the EM Waves.
What are EM waves?Electromagnetic waves can be defined as the waves, that can interact with the matter and are created as a result of vibrations of electric and magnetic fields.Maxwell theoretically predicted the existence of electromagnetic waves.A time varying magnetic field is a source of electric field.Similarly, an oscillating electric field gives rise to an oscillating magnetic field.The amplitudes of electric and magnetic fields can relate as,[tex]E_0=cB_0[/tex]
The expression for intensity of electromagnetic waves in terms of amplitude of magnetic field can be written as,[tex]I=\frac{1}{2u_0} cB_0^{2}[/tex]
How to solve the problem?From the above expression of intensity, we can find the expression for amplitude of magnetic field as,[tex]B_0=\sqrt{\frac{2u_0I}{c} }=\sqrt{\frac{2*4*3.14*10^{-7}*2500 }{3*10^{8} } } =4.57*10^{-6} T[/tex]
Thus, from the amplitude of magnetic field, we get, the amplitude of electric field as,[tex]E_0=cB_0=3*10^{8} *4.57*10^{-6} =1371 V/m[/tex]
Thus, from the above calculations, we get the amplitudes of the oscillating electric and magnetic fields are, [tex]B_0=4.57*10^{-6} T and E_0=1.37*10^{3} V/m[/tex]
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A 43-kg child sits in a massless swing. With what horizontal force must the seat be pulled so that the ropes form an angle of 35o with respect to the vertical? Group of answer choices 300 N 320 N 340 N 360 N
The horizontal force that the seat must be pulled so that the ropes form an angle of 35o with respect to the vertical is 345.2 N.
Horizontal force to be appliedThe horizontal force that the seat must be pulled so that the ropes form an angle of 35o with respect to the vertical is calculated as;
F = Wcosθ
where;
W is weight of the childθ is angle of inclination of the ropeF = (43 x 9.8) x cos(35)
F = 345.2 N
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what provides the centiperal force to a satellite revolving around the earth?
Answer:
the gravitational force
Question 17
The driver of a car traveling along a straight road with a
speed of 72KM ph observes a signboard which give the
speed limit to be 54KM ph. The signboard is 70m ahead
when the driver applies the brakes, calculate the
acceleration of the car which will cause the car to pass the
signboard at the stated speed limit?
So, the acceleration of the car is -1.25 m/s². In other word, the car is also decelerating by 1.25 m/s².
IntroductionHi ! I will help you to discuss about "deceleration in a straight line movement". Please note in advance that deceleration is acceleration which has a negative value. When an object decelerates, the object will continue to move until it reaches a certain speed (which is less than before) or until it stops. The higher the deceleration value, an object that is moving will stop faster and cover a shorter distance.
Formula UsedIn this opportunity, I will give you the following equation to express the relationship between final velocity and initial velocity, acceleration, and distance.
[tex] \boxed{\sf{\bold{(v_t)^2= (v_0)^2 + 2 \times a \times s}}}[/tex]
With the following condition:
[tex] \sf{v_t} [/tex] = final velocity of an object (m/s)[tex] \sf{v_0} [/tex] = initial velocity of an object (m/s)a = acceleration that happen (m/s²)s = the shift or distance of the object (m)Problem SolvingWe know that:
[tex] \sf{v_0} [/tex] = initial velocity of an object = 72 km/h = 20 m/s[tex] \sf{v_t} [/tex] = final velocity of an object = 54 km/h = 15 m/ss = the shift or distance of the object = 70 mNote :
1 m/s = 3.6 km/h. So 10 m/s = 36 km/hWhat was asked ?
a = acceleration that happen = ... m/s²Step by step :
[tex] \sf{(v_t)^2 = (v_0)^2 + 2 \times a \times s} [/tex]
[tex] \sf{15^2 = 20^2 + 2 \times a \times 70} [/tex]
[tex] \sf{225 = 400 + 140 \times a} [/tex]
[tex] \sf{140 a = -175} [/tex]
[tex] \sf{a = \frac{-175}{140}} [/tex]
[tex] \boxed{\sf{\bold{a = -1.25 \: m/s^2}}} [/tex]
ConclusionHere, we see that the acceleration is -1.25 m/s². In other words, the car is also decelerating by 1.25 m/s².
A toy car starts from the rest and accelerates at 1.50m/s2 [E] for 2.25s. What is the final velocity, of the car
3.375m/s is the final velocity of the car.
How do you find final velocity?The final velocity depends on how large the acceleration is and the distance over which it acts.
Initial velocity of an object, you can multiply the acceleration due to a force by the time the force is applied and add it to the initial velocity to get the final velocity.
According to the question,
A toy car starts from the rest and accelerates
So the acceleration = 1.50m/s²
Time = 2.25s
[tex]x=x_{0} + vt[/tex]
[tex]x = 0 + ( 1.50m/s^2*2.25s)[/tex]
[tex]x = 3.375m/s[/tex]
The final velocity, of the car is 3.375 m/s.
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An object on Earth weighs 150 N. What is its mass?
Group of answer choices
15 g
150 kg
15 kg
150 g
Answer:
[tex] \tt \: mass \: = \frac{weight}{acceleration \: due \: to \: gravity} [/tex]
[tex] \longrightarrow \tt \: \frac{150}{10} [/tex]
[tex] \longrightarrow \boxed{ \tt{15 \: kg}}[/tex]
Our final answer is 15 kg .----------- HappY LearninG<3 ------------
You took a second ball headed to a cliff. You dropped this ball straight down with a velocity of 9.3 m/s. The ball hit the ground exactly 8.0s after you released it. How high is the ciff?
Answer:
The cliff is 74.4m.
Explanation:
The reading above clearly state that a ball is dropped straight down with the speed of 9.3m/s and hit the ground exactly at a time of 8.0s after you released it. And it is asking for the distance of the cliff.
In speed formula make distance the subject.
Therefore Distance = speed × time
Distance = 9.3mls × 8.0s = 74.4m
Therefore the cliff is 74.4m.
Determine the launch speed of a horizontally launched projectile that lands 26.3m from the base of a 19.3m high cliff.
The launch velocity of the projectile is 13.28 m/s.
What is projectile motion?
The motion of an object thrown in the air under the force of gravity is known as projectile motion.
Since the object is launched horizontally, its initial velocity along the vertical direction is zero. From the second kinematic equation,
s=u*t+(1/2)at^2.
where s is the displacement, t is the time, u is the initial velocity and a is the acceleration. Since the height is decreasing, so it will be taken negative.
For the vertical motion, s=-19.3 m, a=-9.8 m/s^2 and u=0. Put the values in the above equation and solve it.
-19.3 = (0)*t+(1/2)*(-9.8)*t^2
19.3 = (1/2)*(9.8)*t^2
t=1.98 s
Since the velocity along the horizontal direction is constant, the displacement along the horizontal direction is given by the formula,
X=vt
where X is the horizontal displacement, v is the initial horizontal velocity and t is the time.
For the horizontal motion, X=26.3 m and t=1.98 s. Put the values in this equation and solve it.
26.3=v*(1.98)
v=13.28 m/s
The launch velocity is equal to the initial horizontal velocity, so it is equal to 13.28 m/s.
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