Two inclined planes OA and OB having inclinations `30^@` and `60^@` with the horizontal respectively intersect each other at O, as shown in figure. A particle is projected from point P with velocity `u=10 sqrt(3) m//s` along a direction perpendicular to plane OA. If the particle strikes plane OB perpendicular at Q. Calculate.

(a) time of flight,
(b) velocity with which the particle strikes the plane OB,
(c) height h of point P from point O,
(d) distance PQ. (Take `g=10m//s^(2)`)

Answer

B

Explanation

Let us choose the x and y directions along OB and OA respectively. Then,
` u_x = u = 10(sqrt3) m//s , u_y = 0`
`a_x = -g sin 60^@ = -5 (sqrt3) m//s ^2`
and `a_y = -g cos 60^@ = -5 m//s^2 `
(a) At point Q, x-component of velocity is zero. Hence, substituting in
`v_x = u_x + a_x t `
`0 = 10 sqrt(3) - 5 (sqrt3)t`
or `t = (10 sqrt(3)) / (5sqrt(3)) = 2s`
(b) At point Q, `v = v_y = u_y + a_y t`
`:. v = 0 -(5)(2) = -10 m//s`
Here, negative sign implies that velocity of particles at Q is along negative y-direction.
(c) Distance `PO = |` displacement of particle along y-direction`| = |s_y|`
Here, `s_y = u_yt + 1/2 a_y t^2`
`= 0 - 1/2 (5)(2)^2 = -10 m `
`:. PO = 10m`
Therefore, `h= PO sin 30^@ = (10)(1/2)`
or `h= 5m`.
(d) Distance OQ = displacement of particle along x-direction = `s_x`
Here, `s_x = u_xt + 1/2 a_xt^2`
`= (10 sqrt(3))(2)-1/2 (5sqrt(3))(2)^2 = 10sqrt(3) m`
or `OQ = 10 sqrt(3)m `
`:. PQ = sqrt(PO^2 + (OQ)^2 )`
`=sqrt(10^2+(10sqrt(3))^2)`
`= sqrt(100 + 300) = sqrt(400)`
`:. PQ = 20 m. `