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Physics-

General

Easy

Question

If a stone $s$ to hit at a point which is at a distance $d$ away and at a height $h$ above the point from where the stone starts, then what is the value of initial sped $u$ , if the stone is launched at an angle $Q$ ?

$\frac{g}{\cos θ} \sqrt{\frac{d}{2 (d \tan θ - h)}}$
$\frac{d}{\cos θ} \sqrt{\frac{g}{2 (d \tan θ - h)}}$
$\sqrt{\frac{g d^{2}}{h \cos^{2} θ}}$
$\sqrt{\frac{g d^{2}}{(d - h)}}$

The correct answer is: $fraction numerator d over denominator cos invisible function application theta end fraction square root of fraction numerator g over denominator 2 left parenthesis d tan invisible function application theta minus h right parenthesis end fraction end root$

$h equals open parentheses u sin invisible function application theta close parentheses t minus fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent$
$d equals open parentheses u cos invisible function application theta close parentheses t blank o r blank t equals fraction numerator d over denominator u cos invisible function application theta end fraction$
$h equals u sin invisible function application theta. fraction numerator d over denominator u cos invisible function application theta end fraction minus fraction numerator 1 over denominator 2 end fraction g. fraction numerator d to the power of 2 end exponent over denominator u to the power of 2 end exponent cos to the power of 2 end exponent invisible function application theta end fraction$
$u equals fraction numerator d over denominator cos invisible function application theta end fraction square root of fraction numerator g over denominator 2 left parenthesis d tan invisible function application theta minus h right parenthesis end fraction end root$

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