A ball is projected upwards from a height h a | vedclass

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(C) Let the upward direction be positive and the downward direction be negative. The initial position is $y_0 = h$ and the final position is $y = 0$.T

Vedclass · Accepted Answer

$\frac{v}{g}\left[1+\sqrt{1+\frac{2 g h}{v^2}}\right]$

Vedclass · Answer

(C) Let the upward direction be positive and the downward direction be negative. The initial position is $y_0 = h$ and the final position is $y = 0$.The initial velocity is $u = v$ and the acceleration is $a = -g$.Using the equation of motion $y = y_0 + ut + \frac{1}{2}at^2$,we get:$0 = h + vt - \frac{1}{2}gt^2$Rearranging the terms,we get the quadratic equation:$\frac{1}{2}gt^2 - vt - h = 0$Multiplying by $2$,we get $gt^2 - 2vt - 2h = 0$.Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a=g$,$b=-2v$,and $c=-2h$:$t = \frac{2v \pm \sqrt{(-2v)^2 - 4(g)(-2h)}}{2g}$$t = \frac{2v \pm \sqrt{4v^2 + 8gh}}{2g}$$t = \frac{2v \pm 2\sqrt{v^2 + 2gh}}{2g}$$t = \frac{v \pm \sqrt{v^2 + 2gh}}{g}$Since time must be positive,we take the positive root:$t = \frac{v + \sqrt{v^2 + 2gh}}{g}$$t = \frac{v}{g} + \frac{\sqrt{v^2(1 + \frac{2gh}{v^2})}}{g}$$t = \frac{v}{g} + \frac{v}{g}\sqrt{1 + \frac{2gh}{v^2}}$$t = \frac{v}{g}\left[1 + \sqrt{1 + \frac{2gh}{v^2}}\right]$

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

$A$ ball is projected upwards from a height $h$ above the surface of the earth with velocity $v$. The time at which the ball strikes the ground is

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