Question

A hot-air balloon is rising straight up with a speed of 4.6 m/s. a ballast bag is released from rest relative to the balloon at 13.4 m above the ground. how much time elapses before the ballast bag hits the ground?

Answer 1

The time taken by the ballast bag to reach the ground is 2.18 s

The ballast bag at rest with respect to the balloon has the upward velocity (u) of 4.6 m/s , which is the velocity of the balloon. When it is dropped from the balloon, its motion is similar to an object thrown upwards with an initial velocity u and it falls under the acceleration due to gravity g.

Taking the upward direction as positive and the downward direction as negative, the following equation of motion may be used.

`s=ut+(1)/(2)at^2`

The bag makes a net displacement s of 13.4 m downwards, hence

`s=-13.4 m`

Its initial velocity is

`u=+4.6 m/s`

The acceleration due to gravity acts downwards and hence it is negative.

`g=-9.8 m/s^2`

Use the values in the equation of motion and write an equation for t.

`image`

Solving the equation for t and taking only the positive value for t,

t=2.18 s