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?

WARNING: Rigorous Solution!

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

Here's what we know:

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

`x = 34` `"m"`

`y = 3` `"m"`

and

`alpha_0 = 45^"o"`

We can use the kinematics equations

`Deltay = v_0sinalpha_0t - 1/2g t^2`

and

`Deltax = v_0cosalpha_0t`

(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

• `Deltay` is the change in `y`-position, in this case `3` `"m"`

• `Deltax` is the change in `x`-position, given as `34` `"m"`

• `v_0` is the initial speed (what we're trying to find)

• `alpha_0` is the launch angle, given as `45^"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` `"m/s"^2`

We recognize that the projectile has an `x`-position `3` `"m"` and `y`-position `34` `"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 = (v_0sinalpha_0 +- sqrt((-v_0sinalpha)^2 - 4(g/2)(Deltay)))/g`

Isolating `t` in the second equation:

`t = (Deltax)/(v_0cosalpha_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:

`((Deltax)(secalpha_0))/(v_0) = (v_0sinalpha_0 + sqrt((-v_0sinalpha)^2 - 2g(Deltay)))/g`

(`1/(cosalpha_0) = secalpha_0`)

I eliminated the `+-` 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(v_0sinalpha_0 + sqrt((-v_0sinalpha)^2 - 2g(Deltay))) = (g)(Deltax)(secalpha_0)`
`-------------------`
`v_0(v_0sinalpha_0 + sqrt((v_0)^2sin^2alpha - 2g(Deltay))) = (g)(Deltax)(secalpha_0)`
`-------------------`
`(v_0)^2sinalpha_0 + v_0sqrt((v_0)^2sin^2alpha - 2g(Deltay)) = (g)(Deltax)(secalpha_0)`
`-------------------`
`v_0sqrt((v_0)^2sin^2alpha - 2g(Deltay)) = (g)(Deltax)(secalpha_0) - (v_0)^2sinalpha_0`
`-------------------`
`(v_0)^2((v_0)^2sin^2alpha_0 - 2g(Deltay)) = ((g)(Deltax)(secalpha_0) - (v_0)^2sinalpha_0)^2`
`-------------------`
`(v_0)^4sin^2alpha_0 - (v_0)^2 2g(Deltay) = ((g)(Deltax)(secalpha_0) - (v_0)^2sinalpha_0)^2`
`-------------------`
`(v_0)^4sin^2alpha_0 - (v_0)^2 2g(Deltay) = g^2(Deltax)^2sec^2alpha_0 + (v_0)^4sin^2alpha_0 -2g(Deltax)(v_0)^2 tanalpha_0`
`-------------------`
`-2g(Deltay)(v_0)^2 - g^2(Deltax)^2sec^2alpha_0 + 2g(Deltax)(v_0)^2 tanalpha_0 = 0`
`-------------------`
Collect in terms of `v_0`:

`(v_0)^2(2g(Deltax)tanalpha_0 - 2g(Deltay)) - g^2(Deltax)^2sec^2alpha_0 = 0`
`-------------------`
`(v_0)^2(2g(Deltax)tanalpha_0 - 2g(Deltay)) = g^2(Deltax)^2sec^2alpha_0`
`-------------------`
Divide both sides by `2g(Deltax)tanalpha_0 - 2g(Deltay)`:

`(v_0)^2 = (g^2(Deltax)^2sec^2alpha_0)/(2g(Deltax)tanalpha_0 - 2g(Deltay))`
`-------------------`
`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 = ((sqrt(9.81) (34))/(cos(45^"o")))/(sqrt2 sqrt((34)tan(45^"o") - 3))`

`= 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:

`((Deltax)(secalpha_0))/(v_0) = (v_0sinalpha_0 + sqrt((-v_0sinalpha)^2 - 2g(Deltay)))/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 = (13.5 + sqrt(182 - 58.9))/9.81`

`2.52 = (13.5 + 11.1)/9.81`

`2.52 = 2.51`

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