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

Jan 9, 2016

After $0.088 \text{s}$

#### Explanation:

Use the equation of motion

$v = u + a t$

The vertical component of velocity is given by:

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

So the equation becomes:

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

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

$\theta = \frac{\pi}{3} = {60}^{\circ}$

$\therefore t = 1 \times \sin \frac{60}{9.8} = \frac{0.866}{9.8} = 0.088 \text{s}$