Question

A cart is moving horizontally along a straight line with constant speed 30 m/s. A projectile is to be fired from the moving cart in such a way that it will return to the cart after the cart has moved 80 m. At what speed (Relative to the cart) must the projectiie 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 these

Answer 1

Answer: (C)

As seen from the cart, the projectile the 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 $
For a projectiie going upward
a = - g = - 10 $ms ^{ - 2} $ v = 0
and t = $ \frac{ 8 / 3 }{ 2} = \frac{ 4 }{ 3 } s $
$ \therefore v = u + at $
0 = u - 10 $ \times \frac{ 4 }{ 3} $
u = $ \frac{ 40}{ 3} \, m/s $