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A juggler tosses a ball up in the air with initial speed $u$. At the instant, it reaches its maximum height $H$, he tosses up a second ball with the same initial speed. The two balls will collide at a height
$\frac{H}{4}$
$\frac{H}{2}$
$\frac{3 H}{4}$
$\sqrt{\frac{3}{4}} H$
Solution
(c)
Let first ball reaches upto height $H$ and it fells by a distance $H-h$, where it collided with second ball which rises upto height $h$.
Equation of motion for first and second ball,
$H-h=\frac{1}{2} g t^2 \quad \dots(i)$
$h=u t-\frac{1}{2} g t^2\quad \dots(ii)$
From Eqs. $(i)$ and $(ii)$, we have $H=u t$ or $t=\frac{H}{u}=\frac{u^2 / 2 g}{u}=\frac{u}{2 g}$ Substituting the value of $t$ in Eq. $(ii)$, we have
$h=u \times \frac{u}{2 g}-\frac{1}{2} g \times \frac{u^2}{4 g^2}$
$h=\frac{u^2}{2 g}-\frac{u^2 g}{8 g^2}$
$h=\frac{u^2}{2 g}-\frac{u^2}{8 g}=\frac{4 u^2-u^2}{8 g}$
or $h=3 u^2 / 8 g=\frac{3}{4} \times \frac{u^2}{g}=\frac{3}{4} \cdot H\left[\because H=\frac{u}{2 g}\right]$
