# A projectile is shot from the ground at an angle of (5 pi)/12  and a speed of 5/7 ms^-1. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Sep 1, 2016

In $0.07$ $s$, the projectile will travel $0.013$ $m$ horizontally and $0.024$ $m$ vertically, for a total distance traveled of $0.027$ $m$ from the origin.

#### Explanation:

The initial velocity of the projectile is $\frac{5}{7} = 0.714$ $m {s}^{-} 1$, at an angle of $\frac{5 \pi}{12} = 1.31$ $r a \mathrm{di} a n s$.

This initial velocity is very small, so we are likely to be dealing with small distances and times throughout this calculation. Making this kind of estimation is helpful in checking our answers.

Our first step is to find the horizontal and vertical components of the initial velocity:

${u}_{x} = u \cos \theta = 0.714 \times \cos 1.31 = 0.184$ $m {s}^{-} 1$ (make sure your calculator is in radians mode)

${u}_{y} = u \sin \theta = 0.714 \times \sin 1.31 = 0.690$ $m {s}^{-} 1$

The nature of projectile motion (at least when we simplify it by ignoring air resistance) is that the horizontal velocity, ${u}_{x}$, will remain constant throughout the flight. The vertical velocity, ${u}_{y}$, begins at the value above and decreases to 0 at the top of the trajectory, then becomes increasingly negative until the object returns to ground level.

Our next step is to find the time until the maximum height is reached.

Using $v = u + a t$, rearranged to $t = \frac{v - u}{a}$, and knowing that a is the acceleration due to gravity, which we will write as $- 9.8$ $m {s}^{-} 2$ since it is in the opposite direction to the initial velocity, v is 0 at the top of the trajectory and u is the value of ${u}_{y}$, we get:

$t = \frac{v - u}{a} = \frac{0 - 0.69}{-} 9.8 = 0.07$ $s$

The horizontal distance traveled in this time will be:

$s = {u}_{x} t = 0.184 \times 0.07 = 0.013$ $m$

The vertical distance traveled will be found through:

${v}^{2} = {u}^{2} + 2 a s$, rearranged to $s = \frac{{v}_{y}^{2} - {u}_{y}^{2}}{2 a} = \frac{{0}^{2} - {0.69}^{2}}{2 \times - 9.8} = 0.024$ $m$.

Now that we know how far the projectile has traveled horizontally and vertically, our old friend Pythagoras can give us the total distance traveled:

${a}^{2} = {b}^{2} + {c}^{2}$ rearranged this gives:

$a = \sqrt{{b}^{2} + {c}^{2}} = \sqrt{{0.013}^{2} + {0.024}^{2}} = 0.027$ $m$