Question

A hot-air balloon is ascending at a rate of 7.5m/s when a passenger drops a camera. If the camera is 27m above the ground when its dropped, how much time does it take for the camera to reach the ground? What is its velocity just before it lands?

Answer 1

Answer:

v_{f}=21.75[m/s] (velocity just before it lands)

t = 2.98 [s]

Explanation:

The key to solving this problem is the initial velocity since when the camera falls it is moving up, we will take this velocity as negative and the acceleration gravitation down as positive.

We will use the following equation of kinematics to solve this problem.

$V_{f}^{2} =V_{o}^{2}+2*g*y$

where:

Vf = final velocity [m/s²]

Vo = initial velocity = 7.5 [m/s]

g = gravity acceleration = 9.81 [m/s²]

y = elevation = 27 [m]

$v_{f}^{2} = - (7.5)^{2} +(2*9.81*27)\\ v_{f}^{2} = 473.5\\v_{f}=\sqrt{473.5}\\v_{f}=21.75[m/s]$

Now using the following equation of kinematics, we can calculate the time that the fall lasts.

$v_{f}=v_{o}+g*t$

$21.75 = -7.5 +9.81*t\\9.81*t = 29.25\\t = 2.98[s]$