The height y and the distance x along the horizontal plane of a prohectile on a certain planet (with no surrounding atmosphere) are given by y=8t-`5t^(2)` meter and x=6t meter,where t is in seconds.The velocity with which the projectile is projected is
(Acceleration due to gravity =9.8 m `s^(-2)`)

Options

A. `6ms^(-1)`
B. `8ms^(-1)`
C. `10ms^(-1)`
D. None of these

Answer

C

Explanation

Here `x=6timpliesucosthetaxxt=6t`
`implies ucostheta=6`…(i)
and `2y=16t - 10t^(2)implies y=8t-5t^(2)`
or `usintheta t-1/2g t=8t-5t^(2)`
`impliesusintheta=8`…(ii)
Squaring and adding (i) and (ii) gives u=10m/s