# A projectile is shot from the ground at a velocity of 9 m/s at an angle of pi/3. How long will it take for the projectile to land?

Jun 1, 2016

It takes $1.59$ s.

#### Explanation:

What makes the projectile to land is the gravity. This is a purely vertical component of the motion.
So we are interested in what the projectile does on the vertical axis.

The velocity vector forms an angle of $\frac{\pi}{3}$ with the ground, so it has the vertical component given by

${v}_{v} = 9 \cdot \sin \left(\frac{\pi}{3}\right) = 9 \frac{\sqrt{3}}{2} \setminus \approx 7.79$ m/s.

The equation of motion for an object with free fall is

$s = {s}_{0} + {v}_{0} t + \frac{1}{2} g {t}^{2}$

where ${s}_{0}$ is the initial position, for us it is zero, ${v}_{0}$ is the initial velocity, for us it is ${v}_{v}$, $g$ is the acceleration of gravity $- 9.81 \frac{m}{s} ^ 2$ and $t$ is the time. I use the negative sign for the acceleration because I consider the velocity positive going upward, so the acceleration is negative because it goes downward.

Substituting these values we have

$s = 7.79 t - \frac{1}{2} 9.81 {t}^{2}$

We are interested in the position $s = 0$ because we want to know when the body arrive to the ground (that is the origin of our system of coordinates), so we have

$0 = 7.79 t - - 4.9051 {t}^{2}$.

One solution is $t = 0$ and we expect that at zero time the projectile is still on the ground, then we are interested in the solution with $t \setminus \ne 0$. Because of this, we can divide both sides for $t$ and obtain

$0 = 7.79 - 4.9051 t$

and the time is then

$t = - \frac{7.79}{-} 4.9051 \setminus \approx 1.59$ s.