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) time taken = 2.66 s

b) v = 28.34

Explanation:

given,

rate of descending = 2.3 m/s

height of camera above ground = 41 m              

using equation of motion                      

$h = u t + \dfrac{1}{2}gt^2$              

$41 =2.3t + \dfrac{1}{2}\times 9.8\times t^2$

4.9 t² + 2.3 t - 41 =0                      

t = 2.66 ,-3.13                    

time taken = 2.66 s

b) v² = u² + 2 g h

v² = 0 + 2× 9.8 × 41

v = 28.34