Question

A sandbag is dropped from a height of 150 m, from a hot air balloon that is moving upwards with a velocity of 5.0 ms-1. Ignore air resistance.a) What is the initial velocity of the sandbag?b) How long will the bag take to reach the ground?

Answer 1

At the moment when the sandbag is released from the hot air balloon, it attains the same velocity as that of the balloon.

Hence the initial velocity of the sandbag is $\bf{5 m\,s^{-1}\ upwards.}$

Let the time taken by the sandbag to reach the ground be $\sf{t.}$

• The sandbag is released at a height of $\sf{150\ m,}$ so it has to travel the same height, hence its displacement, $\sf{s=150\ m.}$

• The sandbag falls down due to the acceleration due to gravity, neglecting air resistance. Hence, $\sf{a=g=10\ m\,s^{-2}.}$

Hence by second equation of motion,

$\longrightarrow\sf{s=ut+\dfrac{1}{2}\,at^2}$

We consider downward motion as positive, and upward one, negative.

$\longrightarrow\sf{150=-5t+5t^2}$

$\longrightarrow\sf{5t^2-5t-150=0}$

$\longrightarrow\sf{t^2-t-30=0}$

$\longrightarrow\sf{(t-6)(t+5)=0}$

Since the time is non - negative.,

$\Longrightarrow\underline{\underline{\sf{t}=\bf{6\ s}}}$