The height `y` and the distance `x` along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by `y = (8t - 5t^2) m` and `x = 6tm`, where `t` is in seconds. The velocity with which the projectile is projected at `t = 0` is.

Options

A. 8m/sec
B. 6m/sec
C. 10m/sec
D. Not obtainable from the data

Answer

C

Explanation

`y = (8t -5t^(2)), x = 6t`
`v_(y) = (dy)/(dt) = 8-10 t, v_(x) = (dx)/(dt) = 6`
at t = 0
`u = sqrt(u_(x)^(2) + u_(y)^(2)) = sqrt(6^(2) + 8^(2)) = 10 m//s`