6m/sv^2 = 2gdd = v^2/2g = 33.6^2/2*9.8Then out of this you subtract 50m, the height of the helicopterThen again with this formula you calculate its initial speed.......
Distance down = 57.594 m, Initial velocity = 0 m/s
57.594 = ½ * 9.8 * t^2
Time down = √(57.594 ÷ 4.9) = 3.428 s
Total time = 1.245 + 3.428 = 4.673 seconds
OK
At the instant of its release from the helicopter, the object have the same speed with the helicopter.
Given, h=50 m
v=33.6 m/s
u=0 (initial speed)
we have the relation, v = u + gt
=> t = v/g
=> t = (33.6 m/s) / (9.8 m/s^2)
=> t = 3.429 s
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First you figure out from what height was the object free falling in order to reach 33.6m/s
v^2 = 2gd
d = v^2/2g = 33.6^2/2*9.8
Then out of this you subtract 50m, the height of the helicopter
Then again with this formula you calculate its initial speed.
Good luck
