The height of a helicopter above the ground is given by h = 2.65t3, where h is in meters and t is in seconds. at t = 1.55 s, the helicopter releases a small mailbag. how long after its release does the mailbag reach the ground?

Free fall is the motion of the body such that it is only acted upon by the force of gravity. Gravitational acceleration is normally 10 m/s², for a body in free fall the initial velocity (u) is 0.

Considering an equation of linear motion S = 1/2at² + ut but a=g and u=0

Therefore, S = 1/2gt², h (2.65 t³) = S = 2.65 * 1.55³ = 9.868

t² = 2S/g

t² = 9.868 * 2 / 10

= 1.9736

t = √1.9736

Hence, the time taken will be equivalent to the sqrt of 1.9736 (√1.9736) which is equal to 1.405 seconds.