# A projectile is shot from the ground at a velocity of 5 m/s and at an angle of (2pi)/3. How long will it take for the projectile to land?

Jan 7, 2017

$t \approx 0.88 s$

#### Explanation:

Assuming the launch and landing points are at equal altitudes, we can use a kinematic equation to determine the flight time from the launch of the projectile to its maximum altitude (${v}_{f} = 0$) and multiply by two to get the total flight time.

${v}_{f} = {v}_{i} + {a}_{y} \Delta t$

$\implies \Delta t = \frac{{v}_{f} - {v}_{i}}{a} _ y$

$\implies \Delta t = \frac{- {v}_{i}}{a} _ y$

We know that when an object is in free fall, the acceleration is equal to $- 9.8 \frac{m}{s} ^ 2$ (vertically).

Because the projectile is launched at an angle, we will need to break the velocity up into components. This can be done using basic trigonometry.

Where $v$ is the initial velocity, and ${v}_{x}$ and ${v}_{y}$ are the horizontal and vertical components of the velocity, respectively. We can see that:

${v}_{x} = v \cos \left(\theta\right)$
${v}_{y} = v \sin \left(\theta\right)$

We will only require the $y$ component.

${v}_{y} = 5 \cdot \sin \left(\frac{2 \pi}{3}\right) = \frac{5 \sqrt{3}}{2} \frac{m}{s}$

We can now calculate the rise time of the projectile.

$\Delta t = \frac{- {v}_{i}}{a} _ y$

$\Delta t = \frac{- \frac{5 \sqrt{3}}{2} \frac{m}{s}}{- 9.8 \frac{m}{s} ^ 2}$

$\Delta t = \approx 0.44 s$

The total flight time is then $\approx 0.88 s$