Question

10. Two particles are projected in vertical X - Y plane
from point P and Q as shown in figure. If they
perform motion under gravity, then value of u so
that collision take place during motion
(1) 20 m/s
(2) $ mis
m/s
(3) 40 m/s
(4) 15 m/s
​

Answer 1

Answer:

15 m/s

Explanation:

OP/OQ = 3/4 = k

=> OP = 3k  & OQ= 4k

=> Vertical Distance of P from O = 3kSin53 = 2.4k

=> Vertical Distance of Q from O = 4kSin37 = 2.4k

=> P & Q are in same line

Vertical Speed from P = 20Sin37° = 20*0.6 = 12 m/s

Vertical Speed from P = uSin53° = u*0.8 = 0.8u m/s

Let say after t sec they collide

then Vertical Distance should be equal

12t -(1/2)gt² =  0.8ut - (1/2)gt²

=> u = 12/0.8

=> u = 120/8

=> u = 15

15 m/s

Answer 2

:@amitnrw how you know the distance covered will be

Same for both in vetical direction.