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 = 10/3
som
along a direction perpendicular to plane OA. If the particle strikes
plane OB perpendicular at Q (Take g = 10 m/s). Then
The time of flight from P to Q is :
(A) 5 sec
(B) 2 sec
(C) 1 sec
(D)none of these
The speed with which the particle strikes the plane OB is :
(A) 10 m/s
(B) 20 m/s
(C) 30 m/s
(D)40m/s
The height h of point P from the ground is :
(A) 10✓3m
(B) 10 m
(C) 5 m
(D)20m
The distance PQ is :
(A) 20 m
(B) 10/✓3 m
(C) 10 m
(D) 5m
Answers
Answer:
I think there is some error in the values.
Explanation:
A particle is projected from point P with velocity u = 10/3 m/s
along a direction perpendicular to plane OA.
Then let the velocity of this particle is vx=v=10/3 m/s[ towards x axis]
vy=0
Now, two planes OA and OB having inclinations 30° and 60° with
the horizontal respectively intersect each other at O
Then the acceleration along with x axis,
ax=−gsin60°= -5√3 m/s^2
ay=−gcos60° =−5m/s^2
Therefore, At point Q, x-component of velocity is zero. Hence, substituting ux=vx+axt
0=10/3 −5√3 t
=> t=10/3/5√3 =2√3/3 s
The speed with which the particle strikes the plane OB
u=uy=vy+ayt
u=0−(5) x 2√3/3 =−10 √3/3 m/s
(c) Distance PO= displacement of particle along y-direction = dy
Here, dy=vyt+1/2ayt^2
=0−1/2(5)x (2√3/3)^2
=−10/3m
PO= 10/3m
Therefore, h=POsin30° =(10/3)(1/2)
=> h=5/3m.
(d) Distance OQ = displacement of particle along x-direction = dx
Here, dx=vxt+1/2axt^2
=(10/3 x (2√3/3)−1/2(5√3)(2√3/3)^2
=50√3/9 m
or OQ=50√3/9 m
Therefore,
PQ= √ (PO)2+(OQ)^2
=√ (10/3)^2 +(50√3/9)^2
=√1200/81
∴PQ=√1200/81m