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 $90 \,m$.

$A$. Velocity of the body after $1 \,s$ (in $ms^{-1}$)	$I$. $5$
$B$. Horizontal displacement of the body after $1 \,s$ (in $m$)	$II$. $20$
$C$. Vertical displacement of the body after $1 \,s$ (in $m$)	$III$. $10$
$D$. Vertical velocity of the body after $1 \,s$ (in $ms^{-1}$)	$IV$. $22.4$

$A$ 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 ........ $m$.

Similar Questions

In a sports event,a disc is thrown such that it reaches its maximum range of $80 \ m$. The distance travelled in the first $3 \ s$ is $(g = 10 \ m/s^2)$. (in $m$)

On which factors does the value of the range of a projectile depend? On which factors does the maximum value of the range depend?

$A$ body is projected at $t = 0$ with a velocity $10\,ms^{-1}$ at an angle of $60^\circ$ with the horizontal. The radius of curvature of its trajectory at $t = 1\,s$ is $R$. Neglecting air resistance and taking acceleration due to gravity $g = 10\,ms^{-2}$,the value of $R$ is ........ $m$.

Average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory is: (projection speed = $u$,angle of projection from horizontal = $\theta$)

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