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. 6 m `s^(-1)`
B. 8 m `s^(-1)`
C. 10 m `s^(-1)`
D. 14 m `s^(-1)`

Answer

C

Explanation

Here ,x=6t and y=8t-`5t^(2)`
`therefore v_(x)=(dx)/(dt)=(d)/(dt)`(6t)=6
and `v_(y)=(dy)/(dt)=(d)/(dt)(8t-5t^(2))`=8-10t
t t=0
`v_(x)=u_(x)=6` and `v_(y)=u_(y)=8`
Velocity of projection ,u=`sqrt(u_(x)^(2)+u_(y)^(2))`
`=sqrt((6^(2))+(8)^(2))`
`sqrt(36+64)=10 m s^(-1)`