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
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
$\frac{v}{g} + \frac{{2hg}}{{\sqrt 2 }}$
$\frac{v}{g}\left[ {1 - \sqrt {1 + \frac{{2h}}{g}} } \right]$
$\frac{v}{g}\left[ {1 + \sqrt {1 + \frac{{2gh}}{{{v^2}}}} } \right]$
$\frac{v}{g}\left[ {1 + \sqrt {{v^2} + \frac{{2g}}{h}} } \right]$
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
Since direction of $v$ is opposite to the direction of $g$ and $h$ so from equation of motion
$h = - vt + \frac{1}{2}g{t^2}$
$ \Rightarrow g{t^2} - 2vt - 2h = 0$
$ \Rightarrow t = \frac{{2v \pm \sqrt {4{v^2} + 8gh} }}{{2g}}$
$ \Rightarrow t = \frac{v}{g}\left[ {1 + \sqrt {1 + \frac{{2gh}}{{{v^2}}}} } \right]$
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