Answer:
42.05 m/s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Height (h) = 239 m
Acceleration due to gravity (g) = 3.7 m/s²
Final velocity (v) =?
The velocity with which the camera hits the ground can be obtained as follow:
v² = u² + 2gh
v² = 0² + 2 × 3.7 × 239
v² = 0 + 1768.6
v² = 1768.6
Take the square root of both side
v = √(1768.6)
v = 42.05 m/s
Therefore, the velocity with which the camera hits the ground is 42.05 m/s
