A stunt pilot is attempting to drop a water balloon from a moving airplane onto

Question

A stunt pilot is attempting to drop a water balloon from a moving airplane onto a target on the ground. The plane moves at a speed of 80.6 m/s and a 47 degree above the horizontal when the balloon is released. At the point of release, the plane is at an altitude of 740 m. (a) How far horizontally, measured from a point directly below the plane's initial position, will the balloon travel before striking the ground? 1105 How much time will it take to strike the ground? What constant acceleration relation will allow you to find this time? Knowing the time, how can you find the horizontal distance? m (b) At the point just before balloon strikes the ground, what angle does its velocity make with the horizontal? Give your answer as an angle measured below the horizontal.

Explanation / Answer

at time of release, balloon will have same velocity as plane's velocity.

In vertical:

initial vertical velocity.v0y = 80.6 sin47 = 58.9 m/s

ay = -9.8 m/s^2

vertical displacement, y = 0 - 740 = - 740 m

applying y = v0y t + ay t^2 / 2

- 740 = 58.9t - 4.9 t^2

4.9 t^2 - 58.9 t - 740 = 0

t = 19.7 sec

(A) horizontal distance covered,

x = (80.6 cos47) (19.7) = 1083 m

(b) vfx = v0x = 80.6 cos47 = 55 m/s

vfy = (80.6 sin47) - (9.8 x 19.7) = -134 m/s

theta = tan^-1(134 / 55) = 68 deg

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