Two seconds after projection, a projectile is travelling in a direction inclined at `30^@` to the horizontal. After one more second, it is travelling horizontally. Find the magnitude and direction of its velocity.

Options

A. the velocity of projection is `20 sqrt(3)ms^(-1)`
B. the angle of projection is `30^(@)` with horizontal
C. Both (a) and (b) are correct
D. Both (a) and (b) are wrong

Answer

C

Explanation

`T//2 = 2 +1 = 3s` or `T = 6s`
`:. (2u_(y))/(g) = 6`
`:. U_(y) = 30 ms^(-1)`
Further, `tan 30^(@) = (v_(y))/(v_(x)) = (u_(y)-g t)/(u_(x)) = (30 - 20)/(u_(x))`
or `u_(x) = 10 sqrt(3) ms^(-1)`
or `u = sqrt(u_(x)^(2)+u_(y)^(2)) = 20 sqrt(3) ms^(-1)`
`tan theta = (u_(y))/(u_(x)) = (30)(10sqrt(3)) = 1//sqrt(3)` or `theta = 30^(@)`