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Physics-

General

Easy

Question

A particle $A$ is projected from the ground with an initial velocity of $10 blank m s to the power of negative 1 end exponent$ at an angle of 60 $degree$ with horizontal. From what height $h$ should an another particle $B$ be projected horizontally with velocity $5 m s to the power of negative 1 end exponent$ so that both the particles collide in ground at point $C$ if both are projected simultaneously? $left parenthesis g equals 10 blank m s to the power of negative 2 end exponent right parenthesis$

10 m
30 m
15 m
25 m

The correct answer is: 15 m

Horizontal component of velocity of $A$ is 10 $c o s blank 6 0 degree$ or $5 blank m s to the power of negative 1 end exponent$ which is equal to the velocity of $B$ in horizontal direction. They will collide at $C$ if time of flight of the particles are equal or $t subscript A end subscript equals t subscript B end subscript$
$fraction numerator 2 u sin invisible function application theta over denominator g end fraction equals square root of fraction numerator 2 h over denominator g end fraction end root open parentheses therefore h equals fraction numerator 1 over denominator 2 end fraction g t subscript B end subscript superscript 2 end superscript close parentheses$
or $h equals fraction numerator 2 u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application theta over denominator g end fraction$
$equals fraction numerator 2 open parentheses 10 close parentheses to the power of 2 end exponent open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses to the power of 2 end exponent over denominator 10 end fraction equals 15 blank m$

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