# A projectile is shot from the ground at an angle of pi/12  and a speed of 8 /3 m/s. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Jan 11, 2017

At its maximum height, the projectile will have traveled a horizontal distance of $0.18 m$ from the launch point.

#### Explanation:

We want to calculate the range of the projectile at its maximum altitude.

Begin by breaking the initial velocity into components. Given that ${\vec{v}}_{i} = \frac{8}{3} \frac{m}{s}$ and $\theta = \frac{\pi}{12}$:

Where $\frac{\pi}{12} = {15}^{o}$

Using basic trigonometry, we can see that the perpendicular (vertical, $y$) component of the velocity is opposite the angle and is found using the sine function. Similarly, the parallel (horizontal, $x$) component of the velocity is adjacent to the angle and is found using the cosine function.

• $\sin \left(\theta\right) = \frac{o p p .}{h y p .}$

$\implies \sin \left(\theta\right) = \frac{{v}_{y}}{v}$

• $\cos \left(\theta\right) = \frac{a \mathrm{dj} .}{h y p .}$

$\implies \cos \left(\theta\right) = \frac{{v}_{x}}{v}$

We can rearrange to solve for ${v}_{x}$ and ${v}_{y}$:

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

$\implies {v}_{y} = \left(\frac{8}{3} \frac{m}{s}\right) \sin \left(\frac{\pi}{12}\right)$

$\implies {v}_{y} \approx 0.69 \frac{m}{s}$

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

$\implies {v}_{x} = \left(\frac{8}{3} \frac{m}{s}\right) \cos \left(\frac{\pi}{12}\right)$

$\implies {v}_{x} \approx 2.58 \frac{m}{s}$

At the projectile's maximum altitude, it will have ${v}_{y} = 0$ as it changes direction (from moving upward to downward, i.e. positive to negative direction). We can use this, the value we calculated for the perpendicular component of the initial velocity, and the fact that (ignoring air resistance), a projectile has a vertical acceleration of $- 9.8 \frac{m}{s} ^ 2$ (or $- g$) to calculate how much time has passed when the projectile has reached its maximum altitude using a kinematic equation.

We can then put this value for $t$ into another kinematic equation along with the value we calculated for the parallel component of the initial velocity and the fact that for a projectile (ignoring air resistance), ${\vec{a}}_{x} = 0$.

Calculating $\Delta t$:

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

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

$\Delta t = \frac{0 - 0.69 \frac{m}{s}}{- 9.8 \frac{m}{s} ^ 2} = 0.070 s$

Now we calculate how far the projectile has traveled horizontally (i.e. the range) in this amount of time. Remember that there is no horizontal acceleration.

$\Delta x = {v}_{i} \Delta t + \cancel{\frac{1}{2} a \Delta {t}^{2}}$

$\Delta x = \left(2.58 \frac{m}{s}\right) \left(0.070 s\right) \approx 0.18 m$

$\therefore$At its maximum height, the projectile will have traveled a horizontal distance of $0.18 m$ from the launch point.