A packet is dropped from a stationary helicopter, hovering at a height 'h' from the ground level, and reaches the ground in 12 seconds. Calculate:
1) The value of 'h' (displacement)
2) The final velocity of the packet on reaching ground level
Assume the acceleration due to gravity is [tex]9.8 \, \text{m/s}^2[/tex].
Please show how you derived your answers.
Answer (2)
Use kinematic equations to solve:
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
vf = vo + at
vo = 0
Thus,
vf = at
vf = (9.8)(12) = 117.6 m/s
The height from which the packet is dropped is 705.6 meters, and its final velocity upon reaching the ground is 117.6 m/s.
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