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. `8 m//s`
B. `6 m//s`
C. `10 m//s`
D. `0`

Answer

C

Explanation

`x = 6t`
`v_(x) = (d x)/(d t) = 6`
`y = 8t - 5t^(2)`
`v_(y) = (d y)/(d t) = 8 - 10 t`
`t = 0, v_(x) = 6, v_(y) = 8`
`v = sqrt(v_(x)^(2) + v_(y)^(2)) = 10 m//s`