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Grade 11Mechanics

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revanth

10 Years agoGrade 11

1 Answer

Mallikarjun Maram

10 Years ago

Since the P and Q are colliding at $t = \frac {\sqrt {3}}{2u} d$ , their vertical components of their velocities must be equal. That means

$v sin 60^{o} = u sin \theta$ --- eq 1.

Horizontal components are related such that horizontal distance travelled is d in time $t = \frac {\sqrt {3}}{2u} d$ .

Hence, we have

$(v cos 60^{o} + u cos \theta) * \frac{\sqrt{3}}{2u} d = d$ --- eq 2.

Solve eq 1 and 2 to get $v = \frac {u}{\sqrt{3}}$ .

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