# A projectile is shot from the ground at an angle of ( pi)/3  and a speed of 15 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?

May 21, 2017

I got $\text{9.93 m}$ horizontally from the starting horizontal position, with a maximum height of $\text{8.60 m}$ and a diagonal displacement of $\text{13.14 m}$.

We assume or know:

• Horizontal acceleration is ${\vec{a}}_{x} = 0$, since the projectile is let go after its launch (free horizontal motion).
• Starting position is ${\vec{x}}_{i} = 0 , {\vec{y}}_{i} = 0$, on the ground.

Defining up as positive and right as positive, the coordinates along the parabolic arc are given by the system of equations, assuming no air resistance:

${\vec{y}}_{f} = \frac{1}{2} \vec{g} {t}^{2} + {\vec{v}}_{i y} t + {\cancel{{\vec{y}}_{i}}}^{0}$
${\vec{x}}_{f} = {\cancel{\frac{1}{2} {\vec{a}}_{x} {t}^{2}}}^{0} + {\vec{v}}_{i x} t + {\cancel{{\vec{x}}_{i}}}^{0}$

The "distance from the starting point" is $\Delta \vec{x} \equiv {\vec{x}}_{f}$. For the diagonal distance, then

$\vec{d} = \sqrt{{\left(\Delta \vec{x}\right)}^{2} + {\left(\Delta \vec{y}\right)}^{2}}$. $\text{ "" } \boldsymbol{\left(1\right)}$

After our assumptions, here is what we don't know in red:

$\textcolor{red}{{\vec{y}}_{f}} = \frac{1}{2} \vec{g} {t}^{2} + {\vec{v}}_{i y} t$ $\text{ "" } \boldsymbol{\left(2\right)}$
$\textcolor{red}{{\vec{x}}_{f}} = {\vec{v}}_{i x} t$ $\text{ "" "" "" "" } \boldsymbol{\left(3\right)}$

We can find ${\vec{v}}_{i x}$ and ${\vec{v}}_{i y}$ using trigonometry; since $\cos \theta = \frac{{\vec{v}}_{x}}{\vec{v}}$ and $\sin \theta = \frac{{\vec{v}}_{y}}{\vec{v}}$:

${\vec{v}}_{i x} = \text{15 m/s" cdot cos(pi/3) = "7.5 m/s}$
${\vec{v}}_{i y} = \text{15 m/s} \cdot \sin \left(\frac{\pi}{3}\right) = 7.5 \sqrt{3}$ $\text{m/s}$ $\approx$ $\text{12.99 m/s}$

Recall from Calculus that taking the first derivative of $y = f \left(x\right)$ and setting $\frac{\mathrm{dy}}{\mathrm{dx}}$ equal to $0$ gives the ${x}^{\text{*}}$ at which $y$ is either maximized or minimized.

So, the time ${t}^{\text{*}}$ to get to the maximum height of the concave down parabola can be found by taking the derivative of $\left(2\right)$, and setting $\frac{\mathrm{dv} e c {y}_{f} \left(t\right)}{\mathrm{dt}} = 0$.

$\frac{\mathrm{dv} e c {y}_{f}}{\mathrm{dt}} = {\vec{v}}_{f y} = 0 = \vec{g} t + {\vec{v}}_{i y}$

$\implies {t}^{\text{*" = -(vecv_(iy))/(vecg) = -("12.99 m/s")/(-"9.81 m/s"^2) = "1.32 s}}$

As a result, the final $x$ position from $\left(3\right)$ is:

$\textcolor{b l u e}{{\vec{x}}_{f}} = {\vec{v}}_{i x} {t}^{\text{*}}$

$= \text{7.5 m/s" cdot "1.32 s}$

$=$ $\textcolor{b l u e}{\text{9.93 m}}$

From $\left(2\right)$, we also have enough information to get the maximum height, using the max-height time ${t}^{\text{*}}$:

$\textcolor{b l u e}{{\vec{y}}_{f}} = \frac{1}{2} \left(- \text{9.81 m/s"^2)("1.32 s")^2 + (7.5sqrt3 " m/s")("1.32 s}\right)$

$=$ $\textcolor{b l u e}{\text{8.60 m}}$

Therefore, the diagonal distance traveled at max height is from $\left(1\right)$:

$\textcolor{b l u e}{\vec{d}} = \sqrt{{\left(\Delta \vec{x}\right)}^{2} + {\left(\Delta \vec{y}\right)}^{2}}$

$\equiv \sqrt{{\vec{x}}_{f}^{2} + {\vec{y}}_{f}^{2}}$

$= \sqrt{{\left(\text{9.93 m")^2 + ("8.60 m}\right)}^{2}}$

$\approx$ $\textcolor{b l u e}{\text{13.14 m}}$

Jul 11, 2017

distance = $13.15 \setminus m$

#### Explanation:

For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:

$\left.\begin{matrix}v = u + a t & \text{ where " & s="displacement "(m) \\ s=ut+1/2at^2 & \null & u="initial speed "(ms^-1) \\ s=1/2(u+v)t & \null & v="final speed "(ms^-1) \\ v^2=u^2+2as & \null & a="acceleration "(ms^-2) \\ s=vt-1/2at^2 & \null & t="time } \left(s\right)\end{matrix}\right.$

Horizontal Motion

The projectile will move under constant speed (NB we can still use "suvat" equation with a=0).

The projectile will travel a distance $x$ in time $T$, we must resolve the initial speed in the horizontal direction:

$\left\{\begin{matrix}s = & x & m \\ u = & 15 \cos \left(\frac{\pi}{3}\right) = \frac{15}{2} & m {s}^{-} 1 \\ v = & \text{Not Required} & m {s}^{-} 1 \\ a = & 0 & m {s}^{-} 2 \\ T = & t & s\end{matrix}\right.$

So applying $s = u t + \frac{1}{2} a {t}^{2}$ we get

$x = \frac{15}{2} T$

Vertical Motion

The projectile travels under constant acceleration due to gravity. Its speed will be instantaneous $0$ at the maximum of its trajectory, and we consider the same time interval, $T$. Considering upwards as positive, and again resolving the initial speed (vertically this time):

$\left\{\begin{matrix}s = & y & m \\ u = & 15 \sin \left(\frac{\pi}{3}\right) = \frac{15 \sqrt{3}}{2} & m {s}^{-} 1 \\ v = & 0 & m {s}^{-} 1 \\ a = & - g & m {s}^{-} 2 \\ t = & T & s\end{matrix}\right.$

Applying $v = u + a t$ we have:

$0 = \frac{15 \sqrt{3}}{2} - g T \implies T = \frac{15 \sqrt{3}}{2 g}$

Applying $s = \frac{1}{2} \left(u + v\right) t$ we have:

$y = \frac{1}{2} \left(\frac{15 \sqrt{3}}{2} + 0\right) \cdot \frac{15 \sqrt{3}}{2 g} = \frac{675}{8 g}$

From which we get:

$x = \frac{15}{2} \cdot \frac{15 \sqrt{3}}{2 g} = \frac{225 \sqrt{3}}{4 g}$

Thus we can conclude that the point on the trajectory corresponding to the maximum height occurs at a horizontal distance of $x = \frac{225 \sqrt{3}}{4 g}$ and a vertical height of $y = \frac{675}{8 g}$.

So we calculate the distance, $d$, from the starting point using pythagoras:

$d = \sqrt{{x}^{2} + {y}^{2}}$
$\setminus \setminus = \sqrt{{\left(\frac{225 \sqrt{3}}{4 g}\right)}^{2} + {\left(\frac{675}{8 g}\right)}^{2}}$

Taking $g = 9.8 \setminus \left(m {s}^{- 2}\right)$, we have:

$d = 13.1515 \ldots$

Aug 7, 2017

13.13 m

#### Explanation:

The expression for horizontal range d is :

$\textsf{d = \frac{{v}^{2} \sin 2 \theta}{g}}$

$\therefore$$\textsf{d = \frac{{15}^{2} \sin \left(\frac{2 \pi}{3}\right)}{9.81} = 19.86 \textcolor{w h i t e}{x} m}$

Since we are interested in the horizontal distance when the projectile is at its maximum height we can say that, due to the symmetrical flight path, the horizontal displacement will be half this:

Horizontal displacement = 19.86/2 = sf(color(red)(9.93 m)

To get the height reached we can use:

$\textsf{{v}^{2} = {u}^{2} + 2 a s}$

Taking the vertical component of the motion this becomes:

$\textsf{{v}^{2} = {\left(v \sin \theta\right)}^{2} - 2 g h}$

$\therefore$$\textsf{0 = {\left(15 \sin \left(\frac{\pi}{3}\right)\right)}^{2} - 2 \times 9.81 \times h}$

$\textsf{h = \frac{168.75}{2 \times 9.81} = \textcolor{b l u e}{8.6 \textcolor{w h i t e}{x} m}}$

To get the actual distance D from the starting point we use Pythagoras:

$\textsf{{D}^{2} = {8.6}^{2} + {9.93}^{2}}$

$\textsf{{D}^{2} = 73.97 + 98.60}$

$\textsf{D = 13.13 \textcolor{w h i t e}{x} m}$