Question

A hot-air balloon is descending at a rate of 2.3 m>s when a pas- senger drops a camera. If the camera is 41 m above the ground when it is dropped, (a) how much time does it take for the cam- era to reach the ground, and (b) what is its velocity just before it lands? Let upward be the positive direction for this problem.

Answer 1

Answer:

a) t = 2.64s

b) Vf = -28.7m/s

Explanation:

If the balloon is descending, the velocity is -2.3m/s. So the equation to describe the postion of the falling camera is:

$Y = Vo*t - \frac{g*t^{2}}{2}$

$-41 = -2.3*t - \frac{10*t^{2}}{2}$    Solving for t, we get:

t1 = -3.1s     and      t2 = 2.64s We discard the negative time and use the positive one.

The velocity of the camera will be:

Vf = Vo - g*t = -2.3 - 10*2.64 = -28.7m/s