Question

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2 particles are projected vertically upwards from points (0,0) & (1,0) with uniform velocity 10 m/s & V m/s respectively, as shown in fig. It is found that they collide after time 't'. Find 'v' & 't'.

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Answer 1

Explanation:

$considering \: motion \: in \: \: horizontal \: direction \: \\ time \: taken \: to \: collide \: \\ t = \frac{1}{ \frac{10 \sqrt{3} }{2} - \frac{v}{ \sqrt{2} } } \\ t = \frac{2 \sqrt{2} }{10 \sqrt{6} - 2v} ...1 \\ in \: vertical \: direction \: = 10 \times \frac{1}{2} t - \frac{1}{2}g {t}^{2} \\ \frac{v}{ \sqrt{2} } \times t - \frac{1}{2} g {t}^{2} \\ \\ 5 = \frac{v}{ \sqrt{2} } \\ \\ v = 5 \sqrt{2} \\ \\ t = \frac{2 \sqrt{2} }{5 \sqrt{6} - 10 \sqrt{2} } = 1.49seconds$

hope it helps you out.

Answer 2

Answer:

sorry this is half answer