Question

A cart is moving horizontally along a straight line with constant speed $30{\text{ms}}^{-1}$. 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 $80\text{m}$. At what speed (relative to the cart) must the projectile be fired? (Take $g=10{\text{ms}}^{-1}$)

• A. $10{\text{ms}}^{-1}$
• B. $10\sqrt{8}{\text{ms}}^{-1}$
• C. $\frac{40}{3}{\text{ms}}^{-1}$
• D. None of these

Answer 1

The correct option is C $\frac{40}{3}{\text{ms}}^{-1}$


Let the projectile be fired at speed $v$ vertically w.r.t. cart.
Initial horizontal velocity of projectile
$u_x=u=30\text{m/s}$ and
$u_y=v$

Time taken for vertical journey = Time taken for horizontal journey = Time of flight $\left(T\right)$

$\Rightarrow \frac{2v}{g}=\frac{x}{u_x}=\frac{80}{30}$
$\therefore v=\frac{80\times 10}{30\times 2}=\frac{40}{3}\text{m/s}$