# If a projectile is shot at an angle of pi/6 and at a velocity of 24 m/s, when will it reach its maximum height??

Jun 16, 2016

$t \approx 1.223 \text{ seconds}$ to 2 decimal places

Note that $9.81$ for g is an approximate figure so the solution is an approximation to close limits.

#### Explanation:

Let time in seconds be $t$
Let vertical height by $h$

Vertical height at any time is

$h = \left[24 \sin \left(\frac{\pi}{6}\right) \times t\right] - \left[\frac{1}{2} \times 9.81 \times {t}^{2}\right]$

The $\underline{\text{rate of change of h}}$ will be 0 at maximum height

$\implies \frac{\mathrm{dh}}{\mathrm{dt}} \approx 24 \sin \left(\frac{\pi}{6}\right) - 9.81 t$

But at maximum height $\frac{\mathrm{dy}}{\mathrm{dt}} = 0$ giving

$9.81 t \approx 24 \sin \left(\frac{\pi}{6}\right)$

$\implies t = \frac{24}{9.81} \sin \left(\frac{\pi}{6}\right)$

But $s i m \left(\frac{\pi}{6}\right) = \frac{1}{2}$

$\implies t \approx \frac{12}{9.81}$

$t \approx 1.223 \text{ seconds}$ to 2 decimal places