11.2k views
2 votes
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?

1 Answer

1 vote
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.

User Ben Heymink
by
8.4k points