# If a projectile is shot at an angle of (3pi)/8 and at a velocity of 1 m/s, when will it reach its maximum height?

Dec 23, 2015

After $0.094 \text{s}$

#### Explanation:

The vertical component of the velocity is given by:

${v}_{y} = v \sin \theta$

Use the equation of motion:

$v = u + a t$

This becomes:

$v = u - \text{g} t$

$\therefore 0 = v \sin \theta - \text{g} t$

$\therefore t = \frac{v \sin \theta}{g}$

$2 \pi = {360}^{\circ}$
$\therefore \pi = \frac{360}{2} = {180}^{\circ}$
$\therefore \frac{3 \pi}{8} = \frac{3 \times 180}{8} = {67.5}^{\circ} = \theta$
$\therefore t = \frac{1 \times 0.923}{9.8} = 0.094 \text{s}$