# A projectile is shot from the ground at a velocity of 1 m/s and at an angle of (pi)/2. How long will it take for the projectile to land?

Jan 25, 2016

${v}_{f} = {v}_{i} + a t$
$\frac{\pi}{2}$ is straight up
$0 = 1 - \left(9.81 \times t\right)$ and $t = \frac{1}{9.81} \approx .1 s$

#### Explanation:

graph{-(50x)^2 +5 [-1, 1, 0, 5]}
Observe the arbitrary graph that I provided. It show that the projectile will climb up and come down after it reach maximum height. At maximum height the speed ${v}_{f} = 0$ knowing this write
${v}_{f} = {v}_{i} + a t$ we know that v_i = 1m/s; and a = g = -9.81 m/s^2
Thus $0 = 1 - \left(9.81 \times t\right)$ and $t = \frac{1}{9.81} \approx .1 s$