Q. A cart is moving horizontally along a straight line with constant speed${30\ ms}^{- 1}$. A projectile is to be fired from the moving cart in such a way that it will return to the cart has moved 80 m. At what speed (relative to the cart) must the projectile be fired? $(Take\ g = {10\ ms}^{- 2})$
A) ${10\ ms}^{- 1}$
B) $10\sqrt{8}{ms}^{- 1}$
C) $\frac{40}{3}{ms}^{- 1}$
D) None of the above
Explanation
(c) As seen from the cart the projectile moves vertically upward and comes back.
The time taken by cart to cover 80 m
$= \frac{s}{v} = \frac{80}{30} = \frac{8}{3}s$
$Given,\ u = ?,\ v = 0,\ a = - g = 10{ms}^{- 2}$
(for a projectile going upward)
$and\ t = \frac{8/3}{2} = \frac{4}{3}s$
From first equation of motion
$v = u + at$
$0 = u - 10 \times \frac{4}{3}$
$= \frac{40}{3}{ms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.