# A projectile is shot at a velocity of  5 m/s and an angle of pi/4 . What is the projectile's peak height?

Feb 12, 2016

$0.638 m$

#### Explanation:

Using equations of motion for constant linear acceleration in the vertical direction we get

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

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

$= \frac{{0}^{2} - {\left(5 \sin {45}^{\circ}\right)}^{2}}{2 \times 9.8}$

$= 0.638 m$.