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Physics-

General

Easy

Question

A ball is thrown vertically upwards. It was observed at a height $h$ twice with a time interval $increment t$ . The initial velocity of the ball is

$\sqrt{8 g h + g^{2} {(∆ t)}^{2}}$
$\sqrt{8 g h + {(\frac{g ∆ t}{2})}^{2}}$
$\frac{1}{2} \sqrt{8 g h + g^{2} {(∆ t)}^{2}}$
$\sqrt{8 g h + {4 g}^{2} {(∆ t)}^{2}}$

The correct answer is: $fraction numerator 1 over denominator 2 end fraction square root of 8 g h plus g to the power of 2 end exponent open parentheses increment t close parentheses to the power of 2 end exponent end root$

Let the ball be at height $h$ at time $t$ and time $open parentheses t plus increment t close parentheses.$
Then,
$h equals u t minus fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent$ (i)
and $h equals u open parentheses t plus increment t close parentheses minus fraction numerator 1 over denominator 2 end fraction g open parentheses t minus increment t close parentheses to the power of 2 end exponent$ (ii)
Equating Eqs. (i) and (ii), we get
$t equals fraction numerator 2 u minus g increment t over denominator 2 g end fraction$ (iii)
Substituting Eq. (iii) in Eq. (i), we get
$h equals fraction numerator 4 u to the power of 2 end exponent minus g to the power of 2 end exponent open parentheses increment t close parentheses to the power of 2 end exponent over denominator 8 g end fraction$
$rightwards double arrow blank u equals fraction numerator 1 over denominator 2 end fraction square root of 8 g h plus g to the power of 2 end exponent open parentheses increment t close parentheses to the power of 2 end exponent end root$

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