A body is thrown at angle 30^{circ} to the ho | vedclass

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Question

Horizontal component of velocity along $x$-axis after $1 s$ is

$v _{ x }= u _{ x }+ gt$

$v _{ x }= u \cos 	heta+0 	imes 1$

$v _{ x }=30 \co

Vedclass · Accepted Answer

$10\sqrt 7 $

Vedclass · Answer

Horizontal component of velocity along $x$-axis after $1 s$ is

$v _{ x }= u _{ x }+ gt$

$v _{ x }= u \cos 	heta+0 	imes 1$

$v _{ x }=30 \cos 30^{\circ}$

$v _{ x }=15\sqrt3\; m / s$

Vertical component of the velocity along $y$-axis after $1 s$ is

$v _{ y }= u _{ y }- gt$

$v _{ y }=30 \sin 30^{\circ}-10 	imes 1$

$v _{ y }=15-10$

$v _{ y }=5 m / s$

Velocity of a body after 1 s is

$\overrightarrow{ v }= v _{ x } \hat{ i }+ v _{ y } \hat{ j }$

$\overrightarrow{ v }=15\sqrt3 \hat{ i }+5 \hat{ j }$

Magnitude, $v =|\overrightarrow{ v }|=\sqrt{(15\sqrt3)^{2}+(5)^{2}}$

$v =\sqrt{700}$

$v =10 \sqrt{7}\; m / s$

A body is thrown at angle $30^{\circ}$ to the horizontal with the velocity of $30\; m / s$. After $1\;sec$, its velocity will be (in $m/s$) $\left(g=10\; m / s ^{2}\right)$

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