<span>a. To determine the velocity at which the camera strikes the ground:
v^2 = (v0)^2 + 2ay = 0 + 2ay
v = sqrt{ 2ay }
v = sqrt{ (2)(3.7 m/s^2)(239 m) }
v = 42 m/s
The camera impacts the ground with a speed of 42 m/s.
b. To calculate the duration it takes for the camera to reach the bottom:
y = (1/2) a t^2
t^2 = 2y / a
t = sqrt{ 2y / a }
t = sqrt{ (2)(239 m) / 3.7 m/s^2 }
t = 11.4 seconds
The camera descends for 11.4 seconds before hitting the ground.</span>
