A projectile is shot from the ground at an angle of pi/8  and a speed of 16 m/s. When the projectile is at its maximum height, what will its distance from the starting point be?

Jun 15, 2017

$\text{Distance} = 9.59$ $\text{m}$

Explanation:

We're asked to find the projectile's distance from the launch point when it reaches its maximum height. We therefore need to find its $x$- and $y$-position when it's at this point.

Let's first, for reference later, find its initial velocity components:

v_(0x) = v_0cosalpha = (16"m"/"s")cos(pi/8) = 14.8"m"/"s"

v_(0y) = v_0sinalpha = (16"m"/"s")sin(pi/8) = 6.12"m"/"s"

The point at which the particle is at its highest point is when its instantaneous $y$-velocity is equal to $0$. We can find this $y$-position by the equation

${\left({v}_{y}\right)}^{2} = {\left({v}_{0 y}\right)}^{2} + 2 {a}_{y} \left(\Delta y\right)$

The acceleration is $- g$, which is -9.8"m"/("s"^2). The velocity ${v}_{y}$ is $0$, and the initial $y$-velocity was just found to be $6.12 \text{m"/"s}$. Plugging in these values, and solving for the height $\Delta y$, we have

(0)^2 = (6.12"m"/"s")^2 + 2(-9.8"m"/("s"^2))(Deltay)

19.6"m"/("s"^2)(Deltay) = 37.5("m"^2)/("s"^2)

Deltay = color(red)(1.91 color(red)("m"

Now, with this value, let's calculate the time $t$ when this occurs, so that we can then find out how far in the $x$-direction it is. We can use the formula

${v}_{y} = {v}_{0 y} + {a}_{y} t$

We already know everything here to solve for the time, so let's solve this for $t$:

0 = 6.12"m"/"s" + (-9.8"m"/("s"^2))t

(9.8"m"/("s"^2))t = 6.12"m"/"s"

$t = 0.625$ $\text{s}$

Now, with this time we can use the formula

$\Delta x = {v}_{0 x} t$

to calculate the horizontal distance. We calculated the initial $x$-velocity earlier to be $14.8 \text{m"/"s}$, so

Deltax = (14.8"m"/cancel("s"))(0.625cancel("s")) = color(green)(9.24 $\textcolor{g r e e n}{\text{m}}$

Lastly, we can calculate the total distance $r$ from the launch point using the distance formula:

$r = \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2}} = \sqrt{{\left(9.24 \text{m")^2 + (1.91"m}\right)}^{2}}$

= color(blue)(9.59 color(blue)("m"

Therefore, when the projectile is at its maximum height, it is color(blue)(9.59 meters from its launch point.