A projectile is fired at an angle of `30^(@)` to the horizontal such that the vertical component of its initial velocity is `80m//s`. Its time of fight is `T`. Its velocity at `t=T//4` has a magnitude of nearly.

Options

A. `200 ms^(-1)`
B. `300 ms^(-1)`
C. `100 ms^(-1)`
D. None of these

Answer

D

Explanation

`tan 30^(@) = (u_(y))/(u_(x)) :. (1)/(sqrt(3)) = (80)/(u_(x))`
or `u_(x) = 80 sqrt(3) ms^(-1)`
`T = (2u_(y))/(g) = (2 xx 80)/(10) = 16s`
At `t = (T)/(4) = 4s rArr v_(x) = u_(x) = 80 sqrt(3)ms^(-1)` and `v_(y) = u_(y) +a_(y)t`
or `v_(y) = 80 +(-10) (4) = 40 ms^(-1)`
`:.` Speed `=sqrt((80sqrt(3))^(2)+(40)^(2)) = 144.2 ms^(-1)`