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Question

A cart is moving horizontally along a straight line with constant speed 30 ms−1. A particle is to be fired vertically upwards from the moving cart in such a way that it returns to the cart at the same point from where it was projected after the cart has moved 80 m. At what speed (relative to the must the projectile be fired? (Take g=10 ms−2)

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Solution

The correct option is **C** 403 ms−1

Let the projectile be fired at speed v vertically w.r.t. cart.

Initial horizontal velocity of projectile

ux=u=30 m/s and

uy=v

Time taken for vertical journey = Time taken for horizontal journey = Time of flight (T)

⇒2vg=xux=8030

∴v=80×1030×2=403 m/s

Let the projectile be fired at speed v vertically w.r.t. cart.

Initial horizontal velocity of projectile

ux=u=30 m/s and

uy=v

Time taken for vertical journey = Time taken for horizontal journey = Time of flight (T)

⇒2vg=xux=8030

∴v=80×1030×2=403 m/s

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