# A projectile is shot from the ground at a velocity of 6 m/s and at an angle of (3pi)/4. How long will it take for the projectile to land?

Dec 26, 2015

$0.87 \text{s}$

#### Explanation:

Use the equation of motion:

$v = u + a t$

Considering the vertical component of velocity:

This becomes:

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

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

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

$2 \pi = {360}^{\circ}$

$\therefore \pi = \frac{360}{2} = {180}^{\circ}$

$\therefore \frac{3 \pi}{4} = \frac{3 \times 180}{4} = {135}^{\circ} = \theta$

$t = \frac{6 \sin \left(\frac{3 \pi}{4}\right)}{9.8}$

$t = \frac{6 \times 0.707}{9.8} = 0.433 \text{s}$

This is the time to reach the zenith so the total time of flight will be $2 \times 0.433 = 0.866 \text{s}$