# A projectile is shot at a velocity of  64 m/s and an angle of pi/4 . What is the projectile's peak height?

Jan 16, 2016

$104.49 m$

#### Explanation:

We may consider the horizontal and vertical components separately.

For the y-direction, the acceleration is constant under gravity, hence the equations of motion for constant linear acceleration are valid.

${v}^{2} = {u}^{2} + 2 a x$

$\therefore x = \frac{{v}^{2} - {u}^{2}}{2 a}$

$= \frac{{0}^{2} - {\left(64 \sin \left(\frac{\pi}{4}\right)\right)}^{2}}{2 \times 9.8}$

$= 104.49 m$.

Alternatively, you could also have used calculus by finding an equation for $x \left(t\right)$ and setting its derivative to zero to find its maximum value.