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 75.4 m/s and a 41° above the horizontal when the balloon is released. At the point of release, the plane is at an altitude of 900 m.

Question

(a) How far horizontally, in meters and measured from a point directly below the plane's initial position, will the balloon travel before striking the ground?
(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 in degrees below the horizontal.

Answer 1

Vo = 75.4m/s[41o].
Xo = 75.4*Cos41 = 56.9 m/s.
Yo = 75.4*sin41 = 49.5 m/s.

a. Y^2 = Yo^2 + 2g*h = 49.5^2 + 19.6*900 = 20,090,
Y = 142 m/s. = Final Y-component.
Y = Yo + g*T = 142.
49.5 + 9.8T = 142,
T = 9.44 s. to reach gnd.
d = Xo * T = 56.9 * 9.44 = __ m.

b. Tan A = Y/Xo = 142/56.9.
A = 68.2 degrees.

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Answer 2

a. Correction: Y = Yo + g*T = 142.
-49.5 + 9.8T = 142,
T = 19.5 s. to reach gnd.
d = Xo * T = 56.9 * 19.5 = 1, 112 m.