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Home / Doubts and Solutions/JEE/Physics

CE and DF are two walls of equal height 20m from which two particles A and B of same mass are projected.A is projected horizontally towards left while is projected at an angle 37degree  with velocity 15m/s. If A always sees B to be moving perpendicular to EF then the range of A on ground is 

Asked by m.nilu 3rd July 2018, 6:59 PM
Answered by Expert
Answer:
 The figure is drawn as per my understanding of this question.

Ball-A is projected horizontally towards left. Let u be the projection velocity of ball-A.

Velocity at any time t of ball-A is begin mathsize 12px style v subscript A space equals space u i with rightwards arrow on top space plus space g space t space j with rightwards arrow on top end style ...................(1)
Ball-B is projected with angle α to the horizontal and let v be its projection velocity

Velocity at any time t of ball-B is begin mathsize 12px style v subscript B space equals space v space cos alpha space i with rightwards arrow on top space plus space open parentheses v space sin alpha space minus space g space t close parentheses j with rightwards arrow on top end style...................(2)
" Ball-A sees Ball-B always moving vertical to EF " means relative velocity of ball-B with respect to ball-A is in vertical direction,
i.e., horizontal componenet of relative velocity of ball-B vanishes.
 
relative velocity of ball-B with respect to ball-A, begin mathsize 12px style v subscript B minus space v subscript A space end subscript equals space left parenthesis v space cos alpha space minus space u right parenthesis space i with rightwards arrow on top space plus space open parentheses v space sin alpha space minus space 2 space g space t close parentheses j with rightwards arrow on top end style
hence we have, u = v×cosα ; substituting the given values for v and α, we get u = 15×cos37 ≈ 12 m/s
 
To get the time t for Ball-A to hit ground, we solve the following equation
 
20 = (1/2)×g×t2 = (1/2)×10×t2  ;  we get t = 4 ( for convenience of calculation, i have assumed g = 10 m/s2 )
 
Hence horizontal range of Ball-A = 12×2 = 24 m
Answered by Expert 4th July 2018, 2:41 PM
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