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Physics-

General

Easy

Question

A body is projected vertically upwards. The times corresponding to height $h$ while ascending and while descending are $t subscript 1 end subscript blank a n d blank t subscript 2 end subscript$ respectively. Then the velocity of projection is (g is acceleration due to gravity)

$g \sqrt{t_{1} t_{2}}$
$\frac{g t_{1} t_{2}}{t_{1} + t_{2}}$
$\frac{g \sqrt{t_{1} t_{2}}}{2}$
$\frac{g (t_{1} + t_{2})}{2}$

The correct answer is: $fraction numerator g open parentheses t subscript 1 end subscript plus t subscript 2 end subscript close parentheses over denominator 2 end fraction$

In case of motion under gravity, time taken to go up is equal to the time taken to fall down through the same distance.
$T i m e blank o f blank d e s e n t open parentheses t subscript 2 end subscript close parentheses equals blank t i m e blank o f blank a s c e n t open parentheses t subscript 1 end subscript close parentheses equals fraction numerator u over denominator g end fraction$
$therefore T o t a l blank t i m e blank o f blank f l i g h t blank T equals t subscript 1 end subscript plus t subscript 2 end subscript equals fraction numerator 2 u over denominator g end fraction$
Or $blank u equals fraction numerator g open parentheses t subscript 1 end subscript plus t subscript 2 end subscript close parentheses over denominator 2 end fraction$

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