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.

$s=ut+\frac{1}{2}at^2$

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.

$s=ut+\frac{1}{2} at^2 \\ -13.4=4.6t-\frac{1}{2}(9.8)t^2\\ 4.9t^2-4.6t-13.4=0$

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

t=2.18 s