#### Answer

$\dfrac{3}{4}H$
It has been proved that a projectile reaches three-quarters of its maximum height in half the time needed to reach the maximum height.

#### Work Step by Step

Since, $v=u+at$
and $t=\dfrac{u}{g}$ [ v=0] ...(a)
Since, position-time equation of motion.
we have $s=ut+\dfrac{1}{2}gt^2$
See equation (a).
Thus, $s=u(\frac{u}{g})+\dfrac{1}{2}g(\frac{u}{g})^2$
or, $s=\dfrac{u^2}{2g}=H$ ...(b)
or, $s=u \cdot (\dfrac{u}{2g})-\frac{g}{2}(\dfrac{u}{2g})^2$
Hence, $s=\dfrac{3}{4}H$ [ from part (b)]
It has been proved that a projectile reaches three-quarters of its maximum height in half the time needed to reach the maximum height.