A photographer in a helicopter ascending vertically at a constant rate of 12.0m/s accidentally drops?

When a photographer drops a camera, the camera continues to move upwards for a while with initial speed of 12 m/s, it gradually slows down, reaches the peak and start to fall down in a free fall.

1st part - movement from point of drop to peak

To solve this, first thing is to find peak height of camera.

y = y₀ + v₀²/(2 g)

y = 64 + 12² / (2 * 9.81) = 71.34 m

Time required for camera to slow down from v₀=12m/s to a halt under influence of gravity is:

t₁ = v₀ / g

t₁ = 12 / 9.81 = 1.22 s

2nd part - free fall from height y=71.34m

from

y = g t₂² / 2

t₂ = √(2y/g)

t₂ = √ (2*71.34/9.81) = 3.81 s

Total time the camera will take to reach the ground is:

t = t₁ + t₂ = 1.22 + 3.81 = 5.03 s

Speed when it hits:

v = g t₂

v = 9.81 * 3.81 = 37.38 m/s

That's a shame, because he will have to pay for it.