A military helicopter on a training mission is flying horizontally at a speed of

Question

A military helicopter on a training mission is flying horizontally at a speed of 90.0m/s when it accidentally drops a bomb (fortunately, not armed) at an elevation of 390m . You can ignore air resistance
How much time is required for the bomb to reach the earth?
How far does it travel horizontally while falling?
Find the horizontal component of the bomb's velocity just before it strikes the earth.
Find the vertical component of the bomb's velocity just before it strikes the earth.
If the velocity of the helicopter remains constant, where is the helicopter when the bomb hits the ground?

Explanation / Answer

a) time required to reach earth:

S= 1/2 gt2 , as u =0 [ in vertical direction]

So, t =(2S/g) = 8.917 s

b) total distance travelled horizontally = vel x time = 90.0 x 8.9 = 802.5 meters.

c)horizomtal componenet of velocity,just before striking = same as initial horz vel = 90.0 m/s [as no accln]

d)vertical component = gt = 9.81 x 8.917 = 87.5 m/s

e)if velocity of helicopter is the same, then it will be directly above the bomb, when bomb strikes the ground, because nither the bomb's nor the helicopter's horzontal velocity changes during the flight.

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