# A projectile is shot at an angle of pi/4  and a velocity of  3 m/s. How far away will the projectile land?

Apr 17, 2016

$0.916 m$

#### Explanation:

Find the horizontal component of the velocity, by

${V}_{h} = V \cos \theta = 3 \cos \left(\frac{\pi}{4}\right)$

Knowing that the projectile will travel in an arc, and will have $0$ velocity at the top of this arc (the turning point where it begins to travel down again), you can work out the time for the first half of this arc by

$v = u + a t$

$\therefore \frac{v - u}{a} = t$

$t = \frac{3 \cos \left(\frac{\pi}{4}\right) - 0}{9.8} \approx 0.216 s$

Since this is only half the time, multiply it by $2$ to find the total time the projectile is in the air,

$T = 2 \cdot 0.216 s = 0.432 s$.

Now you can find the distance it travels by simply calculating

$s = v t$
$s = 2.121 m {s}^{-} 1 \cdot 0.432 s \approx 0.916 m$