The height `y` and distance `x` along the horizontal plane of a projectile on a certain planet are given by `x = 6t m` and `y = (8t^(2) - 5t^(2))m`. The velocity with which the projectile is projected is

Options

A. 8m/s
B. 6m/s
C. 10m/s
D. zero

Answer

c

Explanation

`v_(x)=(dx)/(dt)=(d)/(dt)(6t)=6`
`v_(y)=(dy)/(dt)=(d)/(dt)(8t-5t^(2))=8-10t`
At `t=0: u_(x)=v_(x)=6m//s`
`u_(y)=8-0=8m//s`
`therefore u=sqrt(u_(x)^(2)+u_(y)^(2))=sqrt(36+64)=10m//s`