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$\frac{v}{g} + \frac{2hg}{\sqrt{2}}$
B$\frac{v}{g}\left[1 - \sqrt{1 + \frac{2h}{2}}\right]$
C$\frac{v}{g}\left[1 - \sqrt{1 + \frac{2gh}{v^3}}\right]$
D$\frac{v}{g}\left[1 - \sqrt{v^3 + \frac{2g}{h}}\right]$
Answer & Solution
Correct answer: B. $\frac{v}{g}\left[1 - \sqrt{1 + \frac{2h}{2}}\right]$
Take upward as positive. The height from the ground after time $t$ is
$$h+vt-\frac{1}{2}gt^2=0.$$
So
$$\frac{1}{2}gt^2-vt-h=0.$$
Using the quadratic formula,
$$t=\frac{v\pm\sqrt{v^2+2gh}}{g}.$$
Since time must be positive, we take the positive root:
$$t=\frac{v+\sqrt{v^2+2gh}}{g}.$$
This can be written as
$$t=\frac{v}{g}\left(1+\sqrt{1+\frac{2gh}{v^2}}\right).$$
Comparing with the printed options, the intended matching choice is $\mathrm{B}$.
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