Question

A can 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 can) must the projectile be fired? (Takes $=10\,m/{{s}^{2}}$ )

• A. $10\, m/s$
• B. $10\sqrt{8} \, m/s$
• C. $\frac{40}{3} \, m/s$
• D. None of these

Answer 1

Answer: (C)

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

Answer 2

As seen from the cart the projectile moves vertically upwards and comes back.
The time taken by cart to cover $80 \,m$
$=\frac{s}{v}=\frac{80}{30}=\frac{8}{3} s$
Here, $u=?, v=0, a=-g=10 \,m / s ^{2}$
(for a projectile going upwards)
and $t=\frac{8 / 3}{2}=\frac{4}{3} s$
From first equation of motion
$v=u+a t$
$0=u-10 \times \frac{4}{3}$
$\therefore u=\frac{40}{3} \,m / s$