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Question

Two particles A and B are projected simultaneously from the top of two towers of height 10 m and 20 m respectively. Particle A is projected upwards at an angle of 45 with horizontal at a speed of 102 m/s and particle B is projected horizontally at a speed of 10 m/s towards A. What is the distance between towers if particles collide in the air ?Take g=10 m/s2.

A
10.5 m
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B
25 m
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C
20 m
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D
35 m
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Solution

The correct option is C 20 m
The velocity components of A,
(vA)x=102×sin45=10 m/s
(vA)x=102×sin45=10 m/s
Similarly,
The velocity components of B,
(vB)x=10 m/s
(vB)y=0
Relative acceleration aAB=aAaB=g(g)=0
When A collides with B, the displacement of A w.r.t B is,
Sy=+10 m
The velocity of A w.r.t B in y direction is
(vAB)y=(vA)y(vB)y=100
(vAB)y=10 m/s
Time taken to collide will be
t=Sy(vAB)y
t=1010=1 s
Similarly, The velocity of A w.r.t B in y direction is
(vAB)x=(vA)x(vB)x=10(10)
(vAB)x=20 m/s

Hence relative separation in horizontal (Distance between towers) is given by:
d=(vAB)x×t
d=20×1=20 m

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