In figure, find the horizontal velocity `u (in ms^(-1))` of a projectile so that it hits the inclined plane perpendicularly. Given `H = 6.25m`.

Answer

5

Explanation

`v_x = u_x + a_xt`
`rArr 0 = u cos 30^@ - g sin 30^@t`
`rArr t = (usqrt(3))/g`
`S_(y) = u_yt + 1/2 a_yt^2`
`rArr -H cos 30^@ = -u sin 30^@ t`
`-1/2 g cos 30^@t^2`
`rArr -H (sqrt3)/2 = (-u)/2 (usqrt3)/g -1/2gsqrt3/2(u^(2)3)/g^(2)`
`rArr u = sqrt(2gH//5) = sqrt(2 xx 10 xx 6.25//5) = 5ms^(-1)`.