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Q.

A ball A is projected with speed $50 m s^{- 1}$ at an angle $37 °$ with horizontal. At the same time, a second ball B is projected in a direction shown in the figure with speed $60 m s^{- 1}$ from a point 91 m from A. Both ball collide in air. (Given, $\sqrt{3} = 1.7$ ). Then find the time when collision takes place is

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a

1 s

b

2 s

c

$\sqrt{3} s$

d

4 s

answer is A.

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Detailed Solution

Since, both balls are at same levels, so for collision, $v_{A Y} = v_{B Y}$
Question Image
Or $u_{A Y} - g t = u_{B Y} - g t$
$u_{A Y} = u_{B Y}$
$\Rightarrow 50 \sin 37 ° = 60 \sin θ \Rightarrow \sin θ = \frac{50 \times \frac{3}{5}}{60} = \frac{1}{2} \Rightarrow θ = 30 °$
$t = \frac{91}{u_{x r e l}} = \frac{91}{50 \cos 37 ° + 60 \cos θ} = \frac{91}{50 \times \frac{4}{5} + 60 \times \frac{\sqrt{3}}{2}} = \frac{91}{91} = 1 s$

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