# A football is kicked at an angle of 45 degrees from a distance of 34 meters and just clears a 3 meter high goal on its way down. To the nearest m/s what was its initial speed?

Jul 14, 2017

${v}_{0} = 19.1$ $\text{m/s}$

#### Explanation:

WARNING: Rigorous Solution!

We're asked to find the initial speed of a projectile (football) given it was launched at an angle of ${45}^{\text{o}}$ and travels a certain distance.

Here's what we know:

If it just clears the goal, with height of $3$ $\text{m}$ and a horizontal distance of $34$ $\text{m}$, that means we can say at some point along its trajectory, it has position components

$x = 34$ $\text{m}$

$y = 3$ $\text{m}$

and

${\alpha}_{0} = {45}^{\text{o}}$

We can use the kinematics equations

$\Delta y = {v}_{0} \sin {\alpha}_{0} t - \frac{1}{2} g {t}^{2}$

and

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

(I understand this is Precalculus, and you may want an answer involving parametric equations, but here's a Physics approach! These are equations used when an object is treated as a projectile, so don't worry now where these equations come from.)

Parametric equations will not give you the exact answer like this solution will (as terrifying as it is; most of it is just solving for a variable).

where

• $\Delta y$ is the change in $y$-position, in this case $3$ $\text{m}$

• $\Delta x$ is the change in $x$-position, given as $34$ $\text{m}$

• ${v}_{0}$ is the initial speed (what we're trying to find)

• ${\alpha}_{0}$ is the launch angle, given as ${45}^{\text{o}}$

• $t$ is the time (we'll use this variable, even though time is not given)

• $g$ is the acceleration due to gravity near earth's surface, equal to $9.81$ ${\text{m/s}}^{2}$

We recognize that the projectile has an $x$-position $3$ $\text{m}$ and $y$-position $34$ $\text{m}$ at the same time $t$.

Therefore, what we can do is isolate $t$ in one equation, and plug that in for $t$ in the other equation.

Isolating $t$ in the first equation gives

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

Isolating $t$ in the second equation:

$t = \frac{\Delta x}{{v}_{0} \cos {\alpha}_{0}}$

Since $t$ is equal at these positions, we can set these two equations equal to each other, and solve for ${v}_{0}$, the initial speed:

$\frac{\left(\Delta x\right) \left(\sec {\alpha}_{0}\right)}{{v}_{0}} = \frac{{v}_{0} \sin {\alpha}_{0} + \sqrt{{\left(- {v}_{0} \sin \alpha\right)}^{2} - 2 g \left(\Delta y\right)}}{g}$

($\frac{1}{\cos {\alpha}_{0}} = \sec {\alpha}_{0}$)

I eliminated the $\pm$ sign because the football we know is on its way down, so the value for $t$ must be the larger one.

Here's solving for ${v}_{0}$ (skip to bottom if you don't want to see it):

Cross multiply:

${v}_{0} \left({v}_{0} \sin {\alpha}_{0} + \sqrt{{\left(- {v}_{0} \sin \alpha\right)}^{2} - 2 g \left(\Delta y\right)}\right) = \left(g\right) \left(\Delta x\right) \left(\sec {\alpha}_{0}\right)$
$- - - - - - - - - - - - - - - - - - -$
${v}_{0} \left({v}_{0} \sin {\alpha}_{0} + \sqrt{{\left({v}_{0}\right)}^{2} {\sin}^{2} \alpha - 2 g \left(\Delta y\right)}\right) = \left(g\right) \left(\Delta x\right) \left(\sec {\alpha}_{0}\right)$
$- - - - - - - - - - - - - - - - - - -$
${\left({v}_{0}\right)}^{2} \sin {\alpha}_{0} + {v}_{0} \sqrt{{\left({v}_{0}\right)}^{2} {\sin}^{2} \alpha - 2 g \left(\Delta y\right)} = \left(g\right) \left(\Delta x\right) \left(\sec {\alpha}_{0}\right)$
$- - - - - - - - - - - - - - - - - - -$
${v}_{0} \sqrt{{\left({v}_{0}\right)}^{2} {\sin}^{2} \alpha - 2 g \left(\Delta y\right)} = \left(g\right) \left(\Delta x\right) \left(\sec {\alpha}_{0}\right) - {\left({v}_{0}\right)}^{2} \sin {\alpha}_{0}$
$- - - - - - - - - - - - - - - - - - -$
${\left({v}_{0}\right)}^{2} \left({\left({v}_{0}\right)}^{2} {\sin}^{2} {\alpha}_{0} - 2 g \left(\Delta y\right)\right) = {\left(\left(g\right) \left(\Delta x\right) \left(\sec {\alpha}_{0}\right) - {\left({v}_{0}\right)}^{2} \sin {\alpha}_{0}\right)}^{2}$
$- - - - - - - - - - - - - - - - - - -$
${\left({v}_{0}\right)}^{4} {\sin}^{2} {\alpha}_{0} - {\left({v}_{0}\right)}^{2} 2 g \left(\Delta y\right) = {\left(\left(g\right) \left(\Delta x\right) \left(\sec {\alpha}_{0}\right) - {\left({v}_{0}\right)}^{2} \sin {\alpha}_{0}\right)}^{2}$
$- - - - - - - - - - - - - - - - - - -$
${\left({v}_{0}\right)}^{4} {\sin}^{2} {\alpha}_{0} - {\left({v}_{0}\right)}^{2} 2 g \left(\Delta y\right) = {g}^{2} {\left(\Delta x\right)}^{2} {\sec}^{2} {\alpha}_{0} + {\left({v}_{0}\right)}^{4} {\sin}^{2} {\alpha}_{0} - 2 g \left(\Delta x\right) {\left({v}_{0}\right)}^{2} \tan {\alpha}_{0}$
$- - - - - - - - - - - - - - - - - - -$
$- 2 g \left(\Delta y\right) {\left({v}_{0}\right)}^{2} - {g}^{2} {\left(\Delta x\right)}^{2} {\sec}^{2} {\alpha}_{0} + 2 g \left(\Delta x\right) {\left({v}_{0}\right)}^{2} \tan {\alpha}_{0} = 0$
$- - - - - - - - - - - - - - - - - - -$
Collect in terms of ${v}_{0}$:

${\left({v}_{0}\right)}^{2} \left(2 g \left(\Delta x\right) \tan {\alpha}_{0} - 2 g \left(\Delta y\right)\right) - {g}^{2} {\left(\Delta x\right)}^{2} {\sec}^{2} {\alpha}_{0} = 0$
$- - - - - - - - - - - - - - - - - - -$
${\left({v}_{0}\right)}^{2} \left(2 g \left(\Delta x\right) \tan {\alpha}_{0} - 2 g \left(\Delta y\right)\right) = {g}^{2} {\left(\Delta x\right)}^{2} {\sec}^{2} {\alpha}_{0}$
$- - - - - - - - - - - - - - - - - - -$
Divide both sides by $2 g \left(\Delta x\right) \tan {\alpha}_{0} - 2 g \left(\Delta y\right)$:

${\left({v}_{0}\right)}^{2} = \frac{{g}^{2} {\left(\Delta x\right)}^{2} {\sec}^{2} {\alpha}_{0}}{2 g \left(\Delta x\right) \tan {\alpha}_{0} - 2 g \left(\Delta y\right)}$
$- - - - - - - - - - - - - - - - - - -$
color(blue)(v_0 = (sqrt(g) (Deltax)secalpha_0)/(sqrt2 sqrt((Deltax)tanalpha_0 - Deltay))

In all honesty, you could have just plugged in values from the start and solved for ${v}_{0}$, but here is a general equation you can use in any situation if you know the $x$- and $y$-positions and the launch angle ${\alpha}_{0}$.

Plugging in our known values, we have

${v}_{0} = \left(\frac{\sqrt{9.81} \left(34\right)}{\cos \left({45}^{\text{o")))/(sqrt2 sqrt((34)tan(45^"o}}\right) - 3}\right)$

= color(red)(19.1 color(red)("m/s"

It's clearly beautiful:)

We can even check this to make sure our solution is correct, by plugging everything into our original equation:

$\frac{\left(\Delta x\right) \left(\sec {\alpha}_{0}\right)}{{v}_{0}} = \frac{{v}_{0} \sin {\alpha}_{0} + \sqrt{{\left(- {v}_{0} \sin \alpha\right)}^{2} - 2 g \left(\Delta y\right)}}{g}$

(34)/((cos(45^"o"))(19.1)) = ((19.1)sin(45^"o") + sqrt((-(19.1)sin(45^"o"))^2 - 2(9.81)(3)))/9.81

$2.52 = \frac{13.5 + \sqrt{182 - 58.9}}{9.81}$

$2.52 = \frac{13.5 + 11.1}{9.81}$

$2.52 = 2.51$

(slight rounding errors led to this, but it is correct!)