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)

$\frac{ 40 }{ 3 } \, ms^{ - 1}$

Answer 2

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