Q. 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 projectile be fired $($Take$\ g = 10\ m/s^{2})$
A) $10\ m/s$
B) $10\sqrt{8}\ m/s$
C) $\frac{40}{3}\ m/s$
D) None of these
Explanation
(c) As seen from the cart, the projectile moves vertically upward and comes back
The time taken by cart to cover $80\ m$
$\frac{s}{v} = \frac{80}{30} = \frac{8}{3}\ s$
For a projectile going upward , $a = - g = - 10\ m/s^{2},\ \upsilon = 0$
And $t = \frac{8/3}{2} = \frac{4}{3}\ s$
$\therefore\upsilon = u + a\ t \Rightarrow 0 = u - 10 \times \frac{4}{3} \Rightarrow u = \frac{40}{3}\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.