Question

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 24.5 m/s. A 1.0-kg stone is thrown from the basket with an initial velocity of 18.9 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 18.0s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 24.5 m/s.

Part A: How high was the balloon when the rock was thrown out? Express your answer with the appropriate units.

Part B: How high is the balloon when the rock hits the ground? Express your answer with the appropriate units.

Part C: At the instant the rock hits the ground, how far is it from the basket? Express your answer with the appropriate units.

Part D: Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer at rest in the basket. ( |V_h|, |V_v | = ? m/s )

Part E: Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer at rest on the ground. ( |V_h|, |V_v | = ? m/s )

Answer 1

Given,

M = 124 kg ; m = 22 kg ; v = 24.5 m/s ; ms = 1 kg ; vis = 18.9 m/s ; t = 18 s ;

A)from eqn of motion

S = ut + 1/2 at^2

H = 24.5 x 18 + 1/2 x 9.8 x 18^2 = 2028.6 m

Hence, H = 2028.6 m

B)H' = 2028.6 - 24.5 x 18 = 1587.6 m

Hence, H' = 1587.6 m

C)R = 18.9 x 18 = 340.2 m

D = sqrt (1587.6^2 + 340.2^2) = 1623.64 m

Hence, D = 1623.64 m

D)Horizontal remains same

vx = 18.9 m/s

vy = sqrt (2 x 9.8 x 2028.6) - 24.5 = 175 m/s

Hence, vx = 18.9 m/s ; vy = 175 m/s

E)vy = sqrt (2 x 9.8 x 2028.6) = 199.4 m/s

vx = 18.9 m/s ; vy = 199.4 m/s