# Projectile motion: Bi level equation

1. Sep 19, 2009

### Risker

This isn't really a homework question, but rather general interest.
I cam across a curious equation the other day,$$tan\alpha =\frac{a+b}{a}$$

Where alpha is the launch angle of the projectile b is the height of the platform its being shot onto and a is the distance from the launch to the platform.
For example if you where playing golf and trying to hit the ball over a 1m high fence which was 5m away b=1 and a=5.

I was wondering if anyone knew the proof to this equation, i cannot find it anywhere.
It is not exact but it is a very very close approximation.
Thanks everyone.

Last edited: Sep 19, 2009
2. Sep 19, 2009

### Staff: Mentor

A close approximation to what?

I've never seen this and it doesn't seem to make much sense. What is it's purpose? What if a = b? It gives a zero launch angle.

Where did you see it? Do you have a reference?

3. Sep 19, 2009

### Risker

For example if you were playing baseball, and needed to hit it 21.260 Meters. The launch height is 1.3 meters.
using the formula;
b=-1.3 a=21.260
$$\alpha=tan^-1 (\frac{21.260-1.3}{21.260}$$

$$\alpha=43.194 (5sf)$$

This is a very close approximation for the launch angle, which is 43 degrees
A colleague showed me this curious equation, she also had no idea why it worked.

Lets try again; You're launching at 43 degrees and needed a range of 21.260, whats required initial height?
$$b=(a)tan\alpha -a$$
$$b=(21.260)tan(43) -21.260$$
$$b=-1.435$$ (4sf)
Hence height required is 1.435m which is close to the 1.3 proven by traditional means.

If a=b then there is an obvious limitation.

4. Sep 19, 2009

### Staff: Mentor

How did you calculate the launch angle of 43 degrees?

5. Sep 19, 2009

### Risker

I assumed a launch speed of 14ms-1 and used equations of motion.
I shall write it up for you in latex tomorrow morning (10 hours from now, I'm Australian.)

6. Sep 19, 2009

### Staff: Mentor

No need. As long as you realize that the launch angle depends on the speed of the projectile.

7. Sep 19, 2009

### Risker

Thats why it is so strange that the equation works!!!!

8. Sep 19, 2009

### Staff: Mentor

Well, change your assumed speed and see what happens.