## Calculus 8th Edition

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