# Projectile motion question?

## A basketball leaves a player's hands at a height of 2.10 m above the floor. The basket is 2.60 m above the floor. The player likes to shoot the ball at a ${38.0}^{\circ}$ angle. If the shot is made from a horizontal distance of 11.00 m and must be accurate to $\pm 0.22 m$ (horizontally), what is the range of initial speeds allowed to make the basket? Thanks for your help

Jun 30, 2017

${v}_{0} \in \left[10.8 \textcolor{w h i t e}{l} \text{m/s", 11.0color(white)(l)"m/s}\right]$

#### Explanation:

We're asked to find the range of initial speeds that a player can make a shot, with some given measurements.

Let's take a look at our known quantities:

• initial height ${y}_{0}$ is $2.10$ $\text{m}$

• final height $y$ is $2.60$ $\text{m}$

• angle of projection ${\alpha}_{0}$ is ${38.0}^{\text{o}}$

• the minimum horizontal range $\Delta {x}_{\text{min}}$ is $11.00$ $\text{m}$ $- 0.22$ $\text{m}$ $= 10.78$ $\text{m}$

• the maximum horizontal range $\Delta {x}_{\text{max}}$ is $11.00$ $\text{m}$ $+ 0.22$ $\text{m}$ $= 11.22$ $\text{m}$

What we can do to solve this problem is equate two different equations that say when the ball is at a height $2.60$ $\text{m}$ and a horizontal range $10.78$ $\text{m}$.

Using the followming equations, let's solve them for $t$ to eliminate them:

• $x = {x}_{0} + \left({v}_{0} \cos \alpha\right) t$

t = (Deltax)/(v_0cosalpha) = color(purple)((10.78)/(v_0(cos38.0^"o"))

• $y = {y}_{0} + \left({v}_{0} \sin \alpha\right) t - \frac{1}{2} g {t}^{2}$

$t = \frac{{v}_{0} \sin \alpha \pm \sqrt{{\left(- {v}_{0} \sin \alpha\right)}^{2} - 4 \left(\frac{1}{2} g\right) \left(y - {y}_{0}\right)}}{2 \left(\frac{1}{2} g\right)}$

= color(green)((v_0sin38.0^"o" +-sqrt((-v_0sin38.0^"o")^2 - 4(4.9)(2.60-2.10)))/(9.8)

Equating these two equations gives

$\frac{\textcolor{p u r p \le}{\frac{10.78}{{v}_{0} {\left(\cos {38.0}^{\text{o")) = color(green)((v_0sin38.0^"o" +-sqrt((-v_0sin38.0^"o}}\right)}^{2} - 4 \left(4.9\right) \left(2.60 - 2.10\right)}}}{9.8}$

After solving for ${v}_{0}$, you should get a value of

color(red)(10.76 color(red)("m/s"

Now let's find the maximum speed, by changing te horizontal range to $11.22$ $\text{m}$:

$\frac{\textcolor{p u r p \le}{\frac{11.22}{{v}_{0} {\left(\cos {38.0}^{\text{o")) = color(green)((v_0sin38.0^"o" +-sqrt((-v_0sin38.0^"o}}\right)}^{2} - 4 \left(4.9\right) \left(2.60 - 2.10\right)}}}{9.8}$

Solving again for ${v}_{0}$ gives

color(blue)(10.96 color(blue)("m/s"

Therefore, the range of initial speeds is, with three significant figures,

${v}_{0} \in \left[\textcolor{red}{10.8} \textcolor{w h i t e}{l} \textcolor{red}{\text{m/s"), color(blue)(11.0)color(white)(l)color(blue)("m/s}}\right]$

Jul 1, 2017

${u}_{\min} = 10.76 \setminus m {s}^{- 1}$
${u}_{\max} = 10.97 \setminus m {s}^{- 1}$

#### Explanation:

Modelling the basketball as a point projectile and ignoring air resistance.

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 $S$ in time $T$, where $S = 11 \pm 0.22 m$, ie a minimum distance ${s}_{1} = 10.78$ and a maximum distance ${s}_{2} = 11.22$

So we can calculate $s$ using $s = u t$

$S = u \cos \left(38\right) T \implies T = \frac{S}{u \cos 38}$

Vertical Motion

The projectile travels under constant acceleration due to gravity, applied vertically upwards. The projectile must travel a distance $2.6 - 2.1 = 0.5 \setminus m$ vertically.

$\left\{\begin{matrix}s = & 0.5 & m \\ u = & u \sin \left(38\right) & m {s}^{-} 1 \\ v = & \text{Not Required} & m {s}^{-} 1 \\ a = & - g & m {s}^{-} 2 \\ t = & T = \frac{S}{u \cos \left(38\right)} & s\end{matrix}\right.$

Applying $s = u t + \frac{1}{2} a {t}^{2}$ we have:

$0.5 = u \sin \left(38\right) \frac{S}{u \cos \left(38\right)} + \frac{1}{2} \left(- g\right) {\left(\frac{S}{u \cos \left(38\right)}\right)}^{2}$

$\therefore 0.5 = S \tan \left(38\right) - \frac{g {S}^{2}}{2 {u}^{2} {\cos}^{2} \left(38\right)}$

And re-arranging we get:

$\frac{g {S}^{2}}{2 {u}^{2} {\cos}^{2} \left(38\right)} = S \tan \left(38\right) - 0.5$
${u}^{2} = \frac{g {S}^{2}}{2 {\cos}^{2} \left(38\right) \left(S \tan \left(38\right) - 0.5\right)}$

If we take $g = 9.81$, then:

With $S = {S}_{1} = 10.78$ we get:

${u}^{2} = 115.867918 \ldots \implies u = 10.76419 \ldots$

With $S = {S}_{2} = 11.22$ we get:

${u}^{2} = 120.29947 \ldots \implies u = 10.96811 \ldots$

As we used $g$ to $2$ decimal places can only expect our solution to be accurate to $2$ decimal places, thus we have a rage of $u$ from ${u}_{\min} = 10.76$ to ${u}_{\max} = 10.97$