Question

A hot-air balloon is rising straight up with a speed of 4 m/s. A ballast bag is released from rest relative to the balloon at 13 m above the ground. How much time elapses before the ballast bag hits the ground?

Answer 1

Given

velocity of the ballon is u = 4 m/s

height of the ballon when the ballast bag released is h = 13 m

t =?

when the ballast bag is released from rest , with respect to the ground the bag is moving with velocity 4 m/s

and hits the ground after t s

from equations of motion

V^2 - u^2 = 2as

v^2 = u^2+2as

v^2 = 4^2+2*9.8*13

v = 16.45 m/s

for time t

using the equations of motion v = u+at

v = -u+gt

16.45 = -4+9.8*t

solving for t

t = 2.0867 s

so time elapses before the ballast bag hits the ground is 2.0867 s