A projectile is launched at a height of h feet above the ground at an angle of θ with the horizontal. The initial velocity is v0 feet per second, and the path of the projectile is modeled by the parametric equations
x=(v0cosθ)t and y=h+(v0sinθ)t-16t^2
Write the parametric equations to model the path of a projectile launched from ground level (h = 0) at the given values of θ and v0. Use your equations to answer the following questions.

θ=40∘ , v0=128 feet per second
Determine the maximum height
Give the time the object reaches this height (to one decimal place).
Give the approximate horizontal distance the objects is at this point (to one decimal place).
Determine the time the object will hit the ground after it is launched (to one decimal place).

Answers

Answer 1

Using the equation, the time the object will hit the ground after it is launched is 3.4 seconds.

To model the path of a projectile launched from ground level (h = 0), we can set h = 0 in the equation for y:

[tex]y = (v_0sin \theta)t - 16t^2[/tex]

Therefore, the parametric equations for the path of the projectile are:

[tex]x = (v_0cos\theta)t \\y = (v_0sin \theta)t - 16t^2[/tex]

Substituting θ = 40∘ and v0 = 128 feet per second, we get:

[tex]x = (128 cos 40^o)t\\y = (128 sin 40^o)t - 16t^2[/tex]

To determine the maximum height, we need to find the vertex of the parabolic path, which is given by:

t = -b/2a

where a = -16, b = 128 sin 40∘, and c = 0 (since the projectile starts and ends at ground level). Substituting the values, we get:

t = -128 sin 40∘ / (2*(-16)) = 2.3 seconds (approx)

The maximum height can be found by substituting this value of t into the equation for y:

y = (128 sin 40∘)(2.3) - 16(2.3)^2 = 230.3 feet (approx)

Therefore, the maximum height is 230.3 feet and it is reached at 2.3 seconds.

To find the horizontal distance at this point, we can substitute the value of t into the equation for x:

x = (128 cos 40∘)(2.3) = 172.8 feet (approx)

Therefore, the approximate horizontal distance the object is at this point is 172.8 feet.

To determine the time the object will hit the ground after it is launched, we need to find the time when y = 0:

[tex]0 = (128 sin 40)t - 16t^2[/tex]

Solving for t using the quadratic formula, we get:

t = 0 or t = 8 sin 40∘ / 2 = 3.4 seconds (approx)

Therefore, the time the object will hit the ground after it is launched is 3.4 seconds.

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