  Question

# Two inclined planes OA and OB having inclination (with horizontal)  30∘ and 60∘ respectively , intersect each other at O as shown in figure.A particle is projected from point P with velocity along a direction perpendicular to plane OA. if the particle strickes planne OB perpendicularly at Q, calculate , (i) Velocity with which particle strikes the plane OB.   (u) 16.25 (ii) Time of flight    (v) 20 (iii) Verticla height h of P from O (w) 10 (iv) Maximum height from the ground (x)5 (v) Distance PQ  (y)2

A

(i) - (w), (ii) - (y), (iii) - (x), (iv) - (u), (v) - (v)

B

(i) - (x), (ii) - (u), (iii) - (v), (iv) - (w), (v) - (y)

C

(i) - (y), (ii) - (w), (iii) - (x), (iv) - (u), (v) - (v)

D

None of these

Solution

## The correct option is B (i) - (w), (ii) - (y), (iii) - (x), (iv) - (u), (v) - (v) Two given planes are mutually perpendicular and the particle is projected perpendicularly from plane OA. It means →u is parallel to plane OB. At the instant of collision of the particle with OB , its velocity is perpendicular to OB or velocity compoment parallel to OB is zero. For considering motion of particle parallel to plane OB , →u = 10√3ms−1 Acceleration = - g sin  60∘ =−5√3ms−2  = 0 , t = ? S = ? Using v = u + at, t = 2 seconds s = ut +  1−2 a   t−2 or OQ = 10  √3 meter. Now considering motion of the particle normal of the particle normal to plane OB, Intial velocity = 0 , acceleration = h cos 60∘ = 5√3ms−2 T = 2 second , v = ? ,s = PO = ? Using v = u + at, v = 10ms−1 s = ut + 12 a t2 or OP = 10 m h = PO sin 30∘ = 10 * sin 30∘ = 5 m Inclination of →u with the vertical is 30∘, therefore its vertical component is u cos 30∘ = 15ms−1 (upward) Considering vertically upward motion of the particle from P, Intial velocity = 15 m−1 acceleration. = g - 10 10ms−2 v = 0 , S = H = ? Using  v2 =  u2 + 2as , H = 11.25 m. ∴ Maximum height reached by particle by above O = h + H = 16.25 m. Distance PQ = √PQ2+OQ2 = √(10)2+(10√3)2 = 20 m  Suggest corrections   