# Math Help - velocity and angle of projection

1. ## velocity and angle of projection

A ball is projected so that its horizontal range is 45 metres and it passes through a point 22.5 metres horizontally from and 11.25m vertically above the point of projection. Find the angle of projection and the speed of projection. Assume g = 10m/s/s and no air resistance.

I know i should be able to do this, but I can't seem to be able to progress beyond this point in my working.

I have almost zero physics knowledge and trying to approach this from a calculus and simultaneous equations approach.

jacs

2. Originally Posted by jacs
A ball is projected so that its horizontal range is 45 metres and it passes through a point 22.5 metres horizontally from and 11.25m vertically above the point of projection. Find the angle of projection and the speed of projection. Assume g = 10m/s/s and no air resistance.

I know i should be able to do this, but I can't seem to be able to progress beyond this point in my working.

I have almost zero physics knowledge and trying to approach this from a calculus and simultaneous equations approach.

jacs
Who gave you this problem, anyway, and what are they trying to prove by giving it to you??

Okay, we are looking for two equations in the two unknowns v and $\alpha$. Let's first look at the range problem:
$45 = vt~cos(\alpha)$
and
$0 = vt~sin(\alpha) - 5t^2$ <-- y = 0 because the ball is back on the ground.

We need to solve the y equation for t:
$t = \frac{v~sin(\alpha)}{5}$ <-- t = 0 is also a solution. I'm discarding it for obvious reasons.

And we plug that into the x equation:
$45 = v \left ( \frac{v~sin(\alpha)}{5} \right ) ~cos(\alpha)$

$45 = \frac{v^2~sin(\alpha)~cos(\alpha)}{5}$

This is our first equation.

Now let's look at the path of the ball intersecting the point (x, y) = (22.5 m, 11.25 m):
$22.5 = vt~\cos(\alpha)$ <-- This is not the same t as in the range equation!
and
$11.25 = vt~sin(\alpha) - 5t^2$

This time I'm going to solve the x equation for t:
$t = \frac{22.5}{v~cos(\alpha)}$
and insert it into the y equation:
$11.25 = v \left ( \frac{22.5}{v~cos(\alpha)} \right ) ~sin(\alpha) - 5 \left ( \frac{22.5}{v~cos(\alpha)} \right ) ^2$

$11.25 = \frac{22.5~sin(\alpha)}{cos(\alpha)} - \frac{2531.25}{v^2~cos^2(\alpha)}$

This is our second equation.

So we have to solve the simultaneous equations:
$45 = \frac{v^2~sin(\alpha)~cos(\alpha)}{5}$

$11.25 = \frac{22.5~sin(\alpha)}{cos(\alpha)} - \frac{2531.25}{v^2~cos^2(\alpha)}$

There are a variety of ways to attack this system. After staring at it for a while I decided that this was probably the most direct way to do it. Solve the first equation for $v^2$:
$v^2 = \frac{225}{sin(\alpha)~cos(\alpha)}$

and insert this into the bottom equation:
$11.25 = \frac{22.5~sin(\alpha)}{cos(\alpha)} - \frac{2531.25}{\frac{225}{sin(\alpha)~cos(\alpha)} ~cos^2(\alpha)}$

At first glance this looks awful, but simplifying a bit:
$11.25 = \frac{22.5~sin(\alpha)}{cos(\alpha)} - \frac{2531.25}{\frac{225~cos(\alpha)}{sin(\alpha)} }$

$11.25 = \frac{22.5~sin(\alpha)}{cos(\alpha)} - \frac{2531.25~sin(\alpha)}{225~cos(\alpha)}$

$11.25 = 22.5~tan(\alpha) - \frac{2531.25}{225}~tan(\alpha)$
which you can solve for $\alpha$ without any further difficulties.

Let's see how you do from this point on.

-Dan

3. ## Thanks

Thanks soooo much for that. I don't think i ever would have figured that out myself

It was in an assignment i was given, and could do everything else but just that one totally confused me.

Thanks agian.

4. Originally Posted by jacs
Thanks soooo much for that. I don't think i ever would have figured that out myself

It was in an assignment i was given, and could do everything else but just that one totally confused me.

Thanks agian.
No problem. That one was unusually sadistic.

-Dan