# If a projectile is shot at an angle of pi/3 and at a velocity of 26 m/s, when will it reach its maximum height??

Dec 28, 2016

$t = 2.3$ seconds rounded to 1 decimal place

#### Explanation:

The maximum height is when the vertical velocity component of the projectile is equal to the vertically downward velocity due to gravity. Thus we have the model:

$\implies R \sin \left(\theta\right) = \text{g} t$

Where
$R = 26 \frac{m}{s}$
$\theta = \frac{\pi}{3}$
$g \approx 9.81 \frac{m}{s} ^ 2 \leftarrow \text{ known}$

$t$ is time in seconds

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

This becomes:

$\implies 26 \textcolor{red}{\frac{m}{s}} \sin \left(\frac{\pi}{3}\right) = 9.81 \textcolor{red}{\frac{m}{s} ^ 2} \textcolor{w h i t e}{.} t$

You manipulate the units in the same way you do the numbers

$\implies t = \frac{26}{9.81} \sin \left(\frac{\pi}{3}\right) \textcolor{red}{\times \frac{\cancel{m}}{\cancel{s}} \times {s}^{\cancel{\textcolor{w h i t e}{.} 2}} / \left(\cancel{m}\right) \leftarrow \text{ dealing with the units}}$

$t = 2.29527 \ldots$ seconds

$t = 2.3$ seconds rounded to 1 decimal place