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) and at what angle (to the horizontal) must the projectile be fired?

Options

A. `10sqrt(8)` m/s at `45^(@)`
B. `10sqrt(8)` m/s at `cos^(-1)(3)/sqrt(8)`
C. `(40/3)` m/s at `90^(@)`
D. none of these

Answer

C

Explanation

When the cart is moving at 30 m/s, the projectile also has the same horizontal velocity. Also when it is projected at with a velocity u at an angle `theta`, the horizontal component is u cos `theta therefore` Total horizontal velocity is `(u cos theta +30)m//s` But since the projectile travels the same distance as the cart horizontally its horizontal velocity is 30 m/s only. `therefore u cos theta=0` but `u ne0 therefore theta=90^(@)`
`therefore` the projectile is projected at ungle `90^(@)` with horizontal
`thereforet=8/6 s`
at maximum height v=0 `therefore` 0=u+at u=at=`10xx8/6=40/3` m/s