Question

A hot-air balloon is descending at a rate of 1.9 m/s when a passenger drops a camera. If the camera is 47 m above the ground when it is dropped, how much time does it take for the camera to reach the ground? If the camera is 47 m above the ground when it is dropped, what is its velocity just before it lands? Let upward be the positive direction for this problem.

Answer 1

Answer:

The camera lands in t=2.91s with a velocity:

$v=-30.45m/s$

Explanation:

The initial velocity of the camera is the same as the hot-air ballon:

$v_{o}=-1.9m/s$

$y_{o}=47m$

Kinematics equation:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

when the camera lands, y=0:

$0=47-1.9t-4.91*t^{2}$

We solve this equation to find t:

t1=-3.29s       This solution have not sense in our physical point of view

t2=2.91s    

So, the camera lands in t=2.91s

We replace this value in v(t):

$v=v_{o}-g*t=-1.9-9.81*2.91=-30.45m/s$