In figure, the angle of inclination of the inclined plane is `30^@`. Find the horizontal velocity `V_0` so that the particle hits the inclined plane perpendicularly.
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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)`.