Two balls are projected with the same velocit | vedclass

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Question

(b)

As ranges are equal, so the angles of projection of two balls must be complementary.

$	ext { i.e., } 	heta_1+	heta_2 =90^{\circ}$

$	e

Vedclass · Accepted Answer

$3 h$

Vedclass · Answer

(b)

As ranges are equal, so the angles of projection of two balls must be complementary.

$	ext { i.e., } 	heta_1+	heta_2 =90^{\circ}$

$	ext { Given, } 	heta_1 =30^{\circ}$

$	herefore 	heta_2=90^{\circ}-30^{\circ}=60^{\circ}$

Maximum height of projectile (ball),

$H_{\max }=\frac{u^2 \sin ^2 	heta}{2 g}$

$	herefore \quad \frac{H_1}{H_2}=\frac{u_1^2 \sin ^2 	heta_1}{2 g} 	imes \frac{2 g}{u_1^2 \sin ^2 	heta_2}$

$=\frac {\sin ^2\left(30^{\circ}ight)}{\sin ^2\left(60^{\circ}ight)}$ $\left(\because u_1=u_2ight)$

$\frac{h}{H_2}=\frac{\left(\frac{1}{2}ight)^2}{\left(\frac{\sqrt{3}}{2}ight)^2}=\frac{1}{3} \quad\left(\because H_1=hight)$

or $\quad H_2=3 h$

Two balls are projected with the same velocity but with different angles with the horizontal. Their ranges are equal. If the angle of projection of one is $30^{\circ}$ and its maximum height is $h$, then the maximum height of other will be

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