Asked by Sheila
A sandbag is dropped from the helicopter 125 m above the ground. What is its velocity at a height 60.0 m and when it hits the ground?
Answers
Answered by
Damon
Use conservation of energy
Penergy at start = m g h = m*9.81 * 125
Penergy at 60 = m g h = m*9.81*60
so
(1/2) m v^2 = m(9.81)(125-60)
then at the ground
Pe at ground = 0
(1/2) m v^2 = m (9.81)(125)
Penergy at start = m g h = m*9.81 * 125
Penergy at 60 = m g h = m*9.81*60
so
(1/2) m v^2 = m(9.81)(125-60)
then at the ground
Pe at ground = 0
(1/2) m v^2 = m (9.81)(125)
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