# A projectile is shot from the ground at a velocity of 18 m/s and at an angle of (3pi)/4. How long will it take for the projectile to land?

May 29, 2017

$2.6 \text{s}$

#### Explanation:

We're asked to find the time $t$ when the projectile hits the ground with a given initial velocity and launch angle. In terms of physics, we need to find the time $t$ when the height $y$ is zero (there will technically be two times, one time was when the motion first started, and the second time is what we're trying to calculate--when it lands).

To find the time when it lands, we can use the equation

$\Delta y = {v}_{0 y} t - \frac{1}{2} g {t}^{2}$

$\Delta y$ will be $0$ (the change in its height), and we need to find the initial $y$-velocity, ${v}_{0 y}$. We can find this by

${v}_{0 y} = {v}_{0} \sin \alpha = 18 \text{m"/"s"sin((3pi)/4) = color(blue)(12.7"m"/"s}$

Plugging in our known values (g = 9.80"m"/("s"^2)), we have

0 = (12.7"m"/"s")t - 1/2(9.80"m"/("s"^2))t^2

$\left(4.90 \text{m"/("s"^2))t^2 = (12.7"m"/"s}\right) t$

(4.90"m"/("s"^2))t = 12.7"m"/"s"

color(red)(t = 2.6"s"

Thus, the projectile will land after $2.6 \text{s}$.