# A projectile is shot at a velocity of  8 m/s and an angle of pi/6 . What is the projectile's peak height?

Dec 17, 2016

Equations of motion are used to find an answer of 0.816 m

#### Explanation:

Use the equation of motion that reads:

${v}^{2}$ = ${u}^{2}$ - $2 g y$

Where $v$ is the final vertical velocity, $u$ is the initial vertical velocity and $y$ is the change in altitude (the vertical displacement) of the projectile.

Now, the launch angle is $\pi$/6, so $u$ = 8 sin $\frac{\pi}{6}$ = 8 (0.5) = 4.0 $\frac{m}{s}$

"Reaches its greatest height" means $v$ = 0, so putting it together, we get

0 = ${4.0}^{2}$ - 2 (9.8)$y$

$y$ = $\frac{16}{19.6}$ = 0.816 m