A ball is thrown upwards at an angle of 60^o | vedclass

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(C) The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2	heta)}{g}$,where $u$ is the initial velocity,$	heta$ is t

Vedclass · Accepted Answer

$90$

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(C) The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2	heta)}{g}$,where $u$ is the initial velocity,$	heta$ is the angle of projection,and $g$ is the acceleration due to gravity.For the first case,$	heta_1 = 60^o$,so $R_1 = \frac{u^2 \sin(120^o)}{g} = 90 \,m$.For the second case,$	heta_2 = 30^o$,so $R_2 = \frac{u^2 \sin(60^o)}{g}$.Since $\sin(120^o) = \sin(180^o - 60^o) = \sin(60^o)$,it follows that $\sin(2 	imes 60^o) = \sin(2 	imes 30^o)$.Therefore,the range is the same for complementary angles (angles that add up to $90^o$).Since $60^o + 30^o = 90^o$,the range $R_2$ will be equal to $R_1$,which is $90 \,m$.

Column-$I$	Column-$II$
$i)$ The initial velocity of projection	$a)$ $\sqrt{\frac{g(1+a^2)}{2b}}$
$ii)$ The horizontal range of projectile	$b)$ $\frac{a}{b}$
$iii)$ The maximum height attained by projectile	$c)$ $\frac{a^2}{4b}$
$iv)$ The time of flight of projectile	$d)$ $a\sqrt{\frac{2}{bg}}$

$A$ ball is thrown upwards at an angle of $60^o$ to the horizontal. It falls on the ground at a distance of $90 \,m$. If the ball is thrown with the same initial velocity at an angle $30^o$,it will fall on the ground at a distance of ........ $m$.

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