Question

A packet is dropped from a stationary helicopter, hovering at a height 'h' from the ground level, reaches the ground in 12s. Calculate
1) value of 'h' "displacement"
2) final velocity of packet on reaching ground level

take acceleration due to gravity=9.8ms^-2

Please show how you received your answer.

Answer 1

Use kinematic equations to solve:

1) yf = yo + vo*t + 1/2at²

yf = final height
yo = initial height
vo = initial velocity
a = acceleration
t = time

yf - yo = vo*t + 1/2at²

yf - yo = h

vo = 0

Thus,

h = 1/2at²

h = 1/2(9.8)(12)² = 705.6 m

2) vf = vo + at

vo = 0

Thus,

vf = at

vf = (9.8)(12) = 117.6 m/s