# A projectile is shot at an angle of pi/3  and a velocity of  67 m/s. How far away will the projectile land?

Nov 22, 2016

The projectile will land ≈397 meters from its launch point.

#### Explanation:

To determine the range of the projectile, we can break the velocity up into its x- and y-components, then use kinematics to solve for the horizontal distance traveled by the projectile, i.e. its range.

We can start by drawing a diagram of the motion.

Here I have shown two schematic representations. Using basic trigonometry, we can determine the x- and y-components of the velocity, where the initial velocity becomes the hypotenuse of the right triangle (r).

Note: I have converted the angle from radians to degrees.

$\cos \left(\theta\right) = \frac{x}{r}$

$x = r \cos \left(\theta\right)$

$x = 67 \cos \left({60}^{o}\right)$

$x = \frac{67}{2} \frac{m}{s}$

$\sin \left(\theta\right) = \frac{y}{r}$

$y = r \sin \left(\theta\right)$

$y = 67 \sin \left({60}^{o}\right)$

$y = \frac{67 \sqrt{3}}{2}$ $\frac{m}{s}$

We can now use kinematics to solve for the range of the projectile.

We have all of the information necessary to solve for the projectile's flight time, which we can do using y-components of the motion. We can then take this value for $t$ and use it in a kinematic equation with the x-components of the motion to determine the range.

Assuming the projectile has the same initial and final altitude, the rise and fall times of the projectile are equal, i.e. ${t}_{r i s e} = {t}_{f a l l}$, and so the total flight time will simply be twice the rise or fall time.

We can solve for the rise time of the projectile using ${v}_{i} = \frac{67 \sqrt{3}}{2} \frac{m}{s}$, ${v}_{f} = 0$ (the projectile will stop momentarily at its maximum altitude before falling back to Earth), and $a = - g$, where $g$ is the free-fall acceleration on Earth, $9.8 \frac{m}{s} ^ 2$.

v_(fy)=v_(iy)+a_yΔt

Solving for Δt,

Δt=(-v_(iy))/a

Δt=((-67sqrt(3))/2m/s)/(-9.8m/s^2)

Δt=5.92s

The total flight time is equal to twice the rise time, thus

${t}_{f l i g h t} = 11.84 s$.

We can now use this value to solve for the range of the projectile. Because there is no horizontal acceleration in projectile motion, v_x=(Δx)/(Δt) is valid.

Solving for Δx:

Δx=v_xΔt

Δx=(67/2m/s)(11.84s)

Δx=397m