stunt pilot is attempting to drop a water balloon from a mov

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 82.0 m/s and a 35° 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? 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. °

Solution

along horizontal

component of velocity vox = vo*cos35 = 82*cos35 = 67.17 m/s

acceleration ax = 0

along vertical

component of velocity voy = vo*sin35 = 82*sin35 = 47 m/s

acceleration ay = -g

(a)

along vertical

displacement y = -740 m

from equation of motion

y = voy*t + (1/2)*ay*t^2

-740 = 82*sin35*t - (1/2)*9.8*t^2

t = 18 s

along horizontal

x = vox*t + (1/2)*ax*t^2

x = 82*cos35*18

x = 1209 m <<<<<<------------------ANSWER

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(b)

along vertical

velocity before striking vy = voy + ay*t

vy = 82*sin35 - (9.8*18)

vy = -129.4 m/s

along horizontal

vx = vox + ax*t

vx = vox = 82*cos35 = 67.17 m/s

angle = tan^-1(vy/vx)

angle = tan^-1(-129.4/67.17)

angle = 62.5^o measured below the horizontal.