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 $y = 0$.U

Vedclass · Accepted Answer

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

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 $y = 0$.Using the equation of motion $y = y_0 + v_0 t + \frac{1}{2} a t^2$:$0 = h + vt - \frac{1}{2} g t^2$Rearranging the terms to form a quadratic equation in $t$:$\frac{1}{2} g t^2 - vt - h = 0$$g t^2 - 2vt - 2h = 0$Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$t = \frac{2v \pm \sqrt{(-2v)^2 - 4(g)(-2h)}}{2g}$$t = \frac{2v \pm \sqrt{4v^2 + 8gh}}{2g}$Since time must be positive,we take the positive root:$t = \frac{2v + 2\sqrt{v^2 + 2gh}}{2g}$$t = \frac{v}{g} + \frac{\sqrt{v^2(1 + \frac{2gh}{v^2})}}{g}$$t = \frac{v}{g} \left[ 1 + \sqrt{1 + \frac{2gh}{v^2}} \right]$

$A$ 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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