Suppose that a passenger intent on lunch during his first ride in a hot-air balloon accidently drops an apple over the side during the balloon’s liftoff. At the moment of the apple’s release, the balloon is accelerating upward with a magnitude of $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$and has an upward velocity of 2 m / s magnitude. What are the (a) magnitude and (b) direction of the acceleration of the apple just after it is released? (c) Just then, is the apple moving upward or downward, or is it stationery? (d) What is the magnitude of its velocity just then? (e) In the next few moments, does the speed of the apple increase, decrease, or remain constant?

The given data can be listed below,

• The upward acceleration is,$\mathrm{a}=4\mathrm{m}/{\mathrm{s}}^2$
• The upward velocity is, v= 2 m/s.

It deals with the acceleration due to gravity. This gravitation is caused by the gravitational attraction forces.

When the apple is released, it is only under the influence of gravitational force. It is no longer subject to a hot-air balloon. Thus, the acceleration has a magnitude of $\mathrm{a}=9.8\mathrm{m}/{\mathrm{s}}^2$.

Since it is under the influence of gravitational acceleration which acts downward, the direction of acceleration is downward.

When the apple is released, it has an initial upward velocity which is the velocity of a hot-air balloon also. Thus, the apple is moving upward direction.

Since the ball maintains the hot-air balloon’s velocity, the magnitude of the velocity is 2 m/s

Because the acceleration is working downward while the apple is moving upward, the apple's speed will decrease in the coming moments. As a result, the apple's speed will drop until it hits zero, at which point it will begin to increase in the reverse direction.