A helicopter is moving verticallly upwards with a velocity `5 ms^-1`. When the helicopter is at a height `10 m` from ground, a stone is thrown with a velocity `(3 hat i + 4 hat j) m s^-1` from the helicopter w.r.t. the man in it. Considering the point on ground vertically below the helicopter as the origin of coordinates, and the ground below as `xy` plane, find the coordinates of the point where the stone will fall, its distance from origin at the instant the stone strikes the ground, assuming helicopter moves upwards with constant velocity.
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Answer

6 m, 8 m, 10 m

Explanation

Apply `s_z = u_z t + (1)/(2) a_z t^2`
`rArr -10 = 5t - (1)/(2) 10 t^2`
`rArr t^2 - t - 2 = 0`
`rArr t = 2 s`
The stone falls at `P`.
`x = u_x t = 3 xx 2 = 6 m`
`y = u_y t = 4 xx 2 = 8 m`.
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