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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?

User Pugzly
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1 Answer

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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

User Martin Andersson
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