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A cart is moving horizontally along a straight line with a constant speed of 30 ms^-1. A projectile is to be fired from the moving cart in such a way that it will return to the cart (at the same point on cart) after the cart has moved 80 m. At what velocity (relative to the cart) must projectile be fired? (Take, g = 10 ms^-2)

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10ms^-1

$\frac{20}{3} {ms}^{- 1}$

$\frac{40}{3} {ms}^{- 1}$

$\frac{80}{3} {ms}^{- 1}$

answer is C.

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Detailed Solution

The time taken by cart to cover 80 m,

$\frac{s}{v} = \frac{80}{30} = \frac{8}{3} s$

The projectile must be fired (relative to cart) in vertically upward direction.

$i.e. a = - g = - 10 {ms}^{- 2}, v^{'} = 0 and t = \frac{8 / 3}{2} = \frac{4}{3} s$

$∴ v^{'} = u + a t or 0 = u - 10 \times \frac{4}{3} or u = \frac{40}{3} {ms}^{- 1}$

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