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Physics-

General

Easy

Question

A body of mass $m$ is thrown upwards at an angle $theta$ with the horizontal with velocity $v.$ While rising up the velocity of the mass after $t$ second will be

$\sqrt{{(v \cos θ)}^{2} + {(v \sin θ)}^{2}}$
$\sqrt{{(v \cos θ - v \sin θ)}^{2} - g t}$
$\sqrt{v^{2} + g^{2} t^{2} - (2 v \sin θ) g t}$
$\sqrt{v^{2} + g^{2} + g^{2} - (2 v \cos θ) g t}$

The correct answer is: $square root of v to the power of 2 end exponent plus g to the power of 2 end exponent t to the power of 2 end exponent minus open parentheses 2 blank v sin invisible function application theta close parentheses g t end root$

Instantaneous velocity of running mass after $t$ sec will be
$blank v subscript t end subscript equals square root of v subscript x end subscript superscript 2 end superscript plus v subscript y end subscript superscript 2 end superscript end root$
Where, $v subscript x end subscript equals v sin invisible function application theta minus g t equals$ vertical component of velocity,
$v subscript y end subscript equals v cos invisible function application theta equals$ horizontal component of velocity
$v subscript t end subscript equals square root of open parentheses v cos invisible function application theta close parentheses to the power of 2 end exponent plus open parentheses v sin invisible function application theta minus g t close parentheses to the power of 2 end exponent end root$
$v subscript t end subscript equals square root of v to the power of 2 end exponent plus g to the power of 2 end exponent t to the power of 2 end exponent minus 2 v sin invisible function application theta g t end root$

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