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Q.

A ball A is projected with speed 50 ms-1 at an angle 37o with horizontal. At the same time, a second ball B is projected in a direction shown in the figure with speed 60 ms-1 from a point 91 m from A. Both ball collide in air. (Given, 3 = 1.7) The value of θ isThe time when collision takes place is The distance of point of collision from initial position of A is

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a

30∘

b

60∘

c

53∘

d

45∘

e

1 s

f

2 s

g

3 s

h

4 s

i

50 m

j

47.17 m

k

20 m

l

99 m

answer is , , .

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Detailed Solution

Since, both balls are at same levels, so for collision, vAY=vBYor UAY−gt=UBY−gt∴ UAY=UBY⇒ 50sin⁡37∘=60sin⁡θ⇒sin⁡θ=50×3560=12⇒ θ=30∘t=91uxrel =9150cos⁡37∘+60cos⁡θ=9150×45+60×32=9191=1sxA=uAXt=40×1=40mand yA=uAYt−12gt2=30×1−12×10×12=30−5=25∴ sA=xA2+yA2=402+252=47.17m
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