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. `160 ms^(-1)`
C. `144 ms^(-1)`
D. `140 ms^(-1)`

Answer

C

Explanation

Vertical component of initial velocity ,
`u_(y)= usin 30^(@) " or " u = (u_y)/(sin30^(@)) = 80/(1//2)=160 ms^(-1)`
Horizontal component of initial velocity ,
`u_(y)= usin 30^(@) " or " u = (u_y)/(sin30^(@)) = 80/(1//2)=160 ms^(-1)`
`T=(2u sin theta)/g=(2xx160 xx sin30^@)/10 = 16 s`
`:. t = T/4 = 16 /4 = 4 s `
Let v be the velocity of the projectile at `t = T/4`
Its horizontal and vertical components are given by
`v_(x) = u_(x) = 80 sqrt3ms^(-1)`
`v_y=u_y-"gt" =80-10 xx4 = 40 ms^(-1)`
Its magnitude is given by
`v=sqrt(v_x^2+vy^2)=sqrt((80sqrt3)^2+(40)^2)`
`= 40 sqrt(21+1) = 40 sqrt13 ~~ 144 ms^(-1)`