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Question

A cart is moving horizontally along a staight line with a constant speed 30 m/s. A bullet 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 fired, after the cart has moved 80 m. At what speed (relative to the cart), the bullet must be fired ? Take g=10 m/s2.

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Solution

The correct option is **C** 403 m/s

velocity of cart w.r.t ground is −−→vC,G=30 m/s ^i

⇒ Let velocity of bullet with respect to cart is,

−−→vB,C=uy m/s ^j

Hence, velocity of bullet with respect to ground (−−→vB,G) is,

−−→vB,G=−−→vB,C+−−→vC,G

⇒ −−→vB,G=(uy ^j+30 ^i) m/s

−−→aB,G=−g ^j=−10 m/s2 ^j

∵ Fired bullet has velocity component in x & y directions w.r.t ground, and its acceleration is g downwards, hence bullet will follow projectile motion.

⇒Time of flight for bullet will be,

T=2uyg ...(i)

Horizontal distance covered till it is in air,

R=(vB,G)x×T ...(ii)

since bullet returns to cart when it has travelled 80 m

R=80 m ...(iii)

From Eq (i), (ii), (iii),

80=(vB,G)x×2uyg

80=30×2uy10

∴uy=403 m/s

velocity of cart w.r.t ground is −−→vC,G=30 m/s ^i

⇒ Let velocity of bullet with respect to cart is,

−−→vB,C=uy m/s ^j

Hence, velocity of bullet with respect to ground (−−→vB,G) is,

−−→vB,G=−−→vB,C+−−→vC,G

⇒ −−→vB,G=(uy ^j+30 ^i) m/s

−−→aB,G=−g ^j=−10 m/s2 ^j

∵ Fired bullet has velocity component in x & y directions w.r.t ground, and its acceleration is g downwards, hence bullet will follow projectile motion.

⇒Time of flight for bullet will be,

T=2uyg ...(i)

Horizontal distance covered till it is in air,

R=(vB,G)x×T ...(ii)

since bullet returns to cart when it has travelled 80 m

R=80 m ...(iii)

From Eq (i), (ii), (iii),

80=(vB,G)x×2uyg

80=30×2uy10

∴uy=403 m/s

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