A cart is moving horizontally along a straight line with constant speed 30m/s. A projectile is to be fired from the moving a cart is 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) most the projectile be firedd?

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 30m/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`.
`:.` 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 30m/s only.
`:. u cos theta=0" "` but `u!=0" ":.theta=90^(@)`
`:.` the projectile is projected at angle `90^(@)` with horizontal.
Time for journey of cart `=80/30=8/3=2t`
i.e. if t is the time to reach maximum height then the time of flight is `2t=8/3s`
`:.t=8/6s`
at maximum height `v=0:.0=u+at`
`u=-at=10xx8/6=40/3`m/s