A particle is projected horizontally with a speed V=5m/s from the top of a plane inclined at an angle `theta=37^(@)` to the horizontal `(g=10m//s^(2))`

What is the velocity of the particle just before it hits the plane?

Answer

`9.01`

Explanation

`T = ( 2 u sin ( theta + alpha))/( g cos alpha) rArr T = ( 2 xx 5 sin theta )/( 10 cos theta)`
`rArr T = tan theta ` . . . (i)
`R = ( 2 u^(2) sin (theta + 0^(@)) xx cos theta)/( g cos ^(2) theta )`
`R = ( 2 xx 5^(2) xx sin theta xx cos theta )/( 10 cos ^(2) theta )`
`R = ( 5 sin theta )/( 10 xx cos ^(2) theta ) ` . . .(ii)
And `v_(x) = u_(x) + a_(x) t = u cos theta + ( g sin theta ) t`
And ` v_(y) = u sin theta - (g cos theta ) t `
`:. R = ( 5 xx 3// 8)/( 16 // 25) = (17)/(16) m and T = (3)/(4) s `
And at t = T
`v = sqrt((u cos theta + g sin theta xx t )^(2) + ( u sin theta - g cos theta xx t ) ^(2))`
`sqrt(((5 xx 4)/( 5) + (10 xx 3)/( 5) xx (3)/(4)) ^(2) + (( 5 xx 3)/( 5) - (10 xx 4) /(5) xx (3)/( 4))^(2))`
`= ( sqrt( 289 + 36))/( 2) = (5)/(2) sqrt(13) m // s `
= 9.01 m/sec