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(C) Let the first ball reach a maximum height $H = \frac{u^2}{2g}$.At the instant the first ball reaches height $H$,it starts falling. Let the balls c

Vedclass · Accepted Answer

$\frac{3H}{4}$

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(C) Let the first ball reach a maximum height $H = \frac{u^2}{2g}$.At the instant the first ball reaches height $H$,it starts falling. Let the balls collide at a height $h$ from the ground after time $t$ from the moment the second ball is thrown.The first ball falls a distance $(H - h)$ in time $t$. Using the equation of motion for the first ball:$H - h = \frac{1}{2}gt^2 \quad \dots(i)$The second ball rises to height $h$ in time $t$. Using the equation of motion for the second ball:$h = ut - \frac{1}{2}gt^2 \quad \dots(ii)$Adding equations $(i)$ and $(ii)$:$(H - h) + h = \frac{1}{2}gt^2 + ut - \frac{1}{2}gt^2$$H = ut \implies t = \frac{H}{u}$Since $H = \frac{u^2}{2g}$,we have $u = \sqrt{2gH}$. Thus,$t = \frac{H}{\sqrt{2gH}} = \sqrt{\frac{H}{2g}}$.Substituting $t$ into equation $(ii)$:$h = u\left(\frac{H}{u}\right) - \frac{1}{2}g\left(\frac{H}{u}\right)^2 = H - \frac{1}{2}g\left(\frac{H^2}{u^2}\right)$Since $u^2 = 2gH$,we get:$h = H - \frac{1}{2}g\left(\frac{H^2}{2gH}\right) = H - \frac{H}{4} = \frac{3H}{4}$.

Column $I$	Column $II$
$(A)$ Height of particle from ground	$(p)$ $0$
$(B)$ Speed of particle	$(q)$ $10$ $SI$ unit
$(C)$ Displacement of particle	$(r)$ $40$ $SI$ unit
$(D)$ Acceleration of particle	$(s)$ $20$ $SI$ unit

$A$ 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:

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