Answer:
42.1m/s
Explanation:
Given parameters:
Height of cliff = 239m
Free fall acceleration = -3.7m/s²
Unknown:
Final velocity of the camera = ?
Solution:
When the body is falling under free fall, the acceleration = 3.7m/s² ;
The appropriate motion equation is:
V² = U² + 2gH
V = final velocity
U = initial velocity
g = free fall acceleration
H = height of the fall
Insert the parameters and solve;
U = 0
V² = 0² + 2 x 3.7 x 239
V² = 1768.6
V = 42.1m/s
