A model rocket is fired from the roof of a 50 ft tall building as shown in Fig.

Question

A model rocket is fired from the roof of a 50 ft tall building as shown in Fig. P8.1. The height of the rocket is given by y(t) = y(0) + v(0)t - 1/2g t^2 ft where y(t) is the height of the rocket at time t, y(0) = H = 50 ft is the initial height of the rocket, v(0) = 150 ft/s is velocity of the rocket, and g = 32.2 ft/s^2 is the acceleration due to gravity. Find the following: Write the quadratic equation for the height y(t) of the rocket. The velocity v(t) = dy(t)/dt. The acceleration a(t) = dv(t)/dt = d^2y(t)/dt^2. The time required to reach the maximum height as well as the corresponding height y_max. Use your results to sketch y(t). Repeat problem P8-1 if H = 15m, v(0) = 49 m/s and g = 9.8 m/^2.

Explanation / Answer

a)

Y(t) =50+150t-(1/2)*32.2*t2

Y(t) =-16.1t2+150t+50

b)

V(t) =dY(t)/dt =-32.2t+150

c)

a(t) =dY(t)/dt =-32.2 ft/s2

d)

Maximum height is reached when V(t)=0

-32.2t+150=0

t=150/32.2=4.66 s

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