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?

Question

Corbin Bender

Physics

2 February, 21:43

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?

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Answers (1)

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Demarion Curtis · Answer 1

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.