A cart is moving horizontally along a straigh | vedclass

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(C) The time taken by the cart to cover $80\,m$ is given by $t = \frac{d}{v} = \frac{80}{30} = \frac{8}{3}\,s$.For the projectile to return to the sam

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$\frac{40}{3}\,m/s$

Vedclass · Answer

(C) The time taken by the cart to cover $80\,m$ is given by $t = \frac{d}{v} = \frac{80}{30} = \frac{8}{3}\,s$.For the projectile to return to the same point on the cart,it must be fired vertically relative to the cart. The total time of flight $T$ must be equal to the time taken by the cart to travel $80\,m$,so $T = \frac{8}{3}\,s$.The time taken to reach the maximum height is $t_h = \frac{T}{2} = \frac{8/3}{2} = \frac{4}{3}\,s$.Using the equation of motion $v = u + at$,where $v = 0$ at maximum height,$a = -g = -10\,m/s^2$,and $t = \frac{4}{3}\,s$:$0 = u - 10 	imes \frac{4}{3}$$u = \frac{40}{3}\,m/s$.Thus,the projectile must be fired with a velocity of $\frac{40}{3}\,m/s$ relative to the cart.

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