A helicopter is moving vertically upwards with a velocity 10 m/s. When the helicopter is at a height of 15 m from ground, a stone is thrown with a velocity ( 6 i ^ + 8 j ^ ) m / s from the helicopte

Question

A helicopter is moving vertically upwards with a velocity 10 m/s. When the helicopter is at a height of 15 m from ground, a stone is thrown with a velocity ( 6 i ^ + 8 j ^ ​ ) m / s 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, if distance (in m ) between the helicopter \& the stone, at the instant the stone strikes the ground is d 3.25 ​ , find the value of d. (Assume helicopter moves upwards with constant velocity).

TestHub · Accepted Answer

For stone ; u z ​ = − 10 m / s so 15 = u z ​ t + 2 1 ​ gt 2 ⇒ t = 3 sec so distance of striking point from origin (in x-y plane) = ∣3 ( 6 i ^ + 8 j ^ ​ ) ∣ = 30 m In 3 sec , distance travelled by helicopter = 10 × 3 = 30 m so it's distance from origin, z = 30 + 15 = 45 m so distance between the helicopter and point of strike at the time of strike = 3 0 2 + 4 5 2 ​ = 30 3.25 ​ m

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