A particle is projected horizontally with speed u from the top of a tower of height h. At the same instant, a target directly below the point of projection starts moving horizontally from rest in the same direction with constant acceleration a. For the projectile to hit the target, a must be:
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Question

A particle is projected horizontally with speed u from the top of a tower of height h. At the same instant, a target directly below the point of projection starts moving horizontally from rest in the same direction with constant acceleration a. For the projectile to hit the target, a must be:

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G1 Physics · Accepted Answer

The projectile reaches the ground after T = √(2h/g). Its horizontal displacement is x p x_p x p ​ = uT. The target starts from rest, so its displacement is x t x_t x t ​ = ½aT². For collision, x t x_t x t ​ = x p . x_p. x p ​ . Thus, ½aT² = uT ⇒ a = 2u/T. Using T = √(2h/g), a = 2u√(g/2h) = u√(2g/h).