# A projectile is shot from the ground at a velocity of 7 m/s at an angle of pi/4. How long will it take for the projectile to land?

Oct 11, 2017

$t = \frac{7 \sqrt{2}}{20} s$

#### Explanation:

to find the time of flight we use the vertical components

so vertically

${u}_{y} = 7 \sin \left(\frac{\pi}{4}\right) = \frac{7 \sqrt{2}}{2} m {s}^{- 1}$

${s}_{y} = 0 m$

$g = - 10 m {s}^{- 2}$

vertically

${s}_{y} = {u}_{y} t + \frac{1}{2} g {t}^{2}$

$\implies 0 = \frac{7 \sqrt{2}}{2} t - 10 {t}^{2}$

$0 = t \left(\frac{7 \sqrt{2}}{2} - 10 t\right)$

$t = 0 \text{ }$this is the initial point of the trajectory

$\therefore \frac{7 \sqrt{2}}{2} - 10 t = 0$

$t = \frac{7 \sqrt{2}}{20} s$