# Homework Help: How to make a proof for complementary angles of elevation.

1. Aug 1, 2012

### Shanyn

1. The problem statement, all variables and given/known data
I have an assignment that is to prove for instance the range of a projectile that is fired from 40 degrees will be the same as one fired from 50 degrees cause they are complementary but I need a general way to prove it for all angles less than 90 degrees.

2. Relevant equations
sin θ = cos 90 - θ

3. The attempt at a solution
I have searched the Internet and read my physics book and I under stand how to work out if they have the same range or not I'm just having trouble coming up with a general proof for it.

2. Aug 1, 2012

### Curious3141

The first thing you should start with is to find an expression relating the range R to the angle θ.

Do you already have an expression, and can you assume it? If not, you have to derive it.

3. Aug 1, 2012

### Shanyn

Ok I'll try.
No I don't have an expression. I really don't have anything other than what's in my question is all I know really.

4. Aug 1, 2012

### Curious3141

Write down the expressions for vertical and horizontal displacement at time t. What should time t be when the particle hits the ground?

5. Aug 1, 2012

### Shanyn

So the vertical = (u sin theta) t
Horizontal = (u cos theta) t -4.9t^2
Depending on the angle would depend on what time is when it hits the ground, wouldn't it.?

6. Aug 1, 2012

### azizlwl

Vertical displacement is zero for the range.

7. Aug 1, 2012

### Shanyn

Ok umm so should
R = (Vx) (t) With Vx= (Vcos theta) and t= (2*Vsin theta)/g
So R should = (-V^2 * sin2 theta)/g
Is that part right?

8. Aug 1, 2012

### voko

Why the minus sign?

9. Aug 1, 2012

### Shanyn

Oops that's not suppose to be there.
Do you have any suggestions on how I could go about proving that that equation is the same as-
R= (V^2 * sin 2* 90- theta ) / g
Also is it suppose to be on gravity or on acceleration?

Thanks heaps for everyone's help :)

10. Aug 1, 2012

### voko

R= (V^2 * sin 2* 90- theta ) / g is not right and it does not follow from (V^2 * sin2 theta)/g

Use brackets.

11. Aug 1, 2012

### Shanyn

R= ( V^2 *sin (2* (90- theta)) / g. << is that what you mean by use brackets or is it wrong altogether? :)

And if it is right if the angles are complementary that R= ( V^2 *sin (2* (90- theta)) / g and R= ( V^2 *sin (2* theta) / g should equal the same range shouldn't they? Or am I completely on the wrong track.? :)

12. Aug 1, 2012

### voko

Recall the unit circle trigonometry and think what sin 2(90 - theta) simplifies to.

13. Aug 1, 2012

### azizlwl

Wrong.
R=Vxt,
t=R/Vx ........(1)

Where you vertical motion equation?
H=? ..........(2)

Subtitute (1) in (2) you get the range R.

14. Aug 1, 2012

### Curious3141

Why is your gravity term in the horizontal displacement equation? Shouldn't it be in the vertical equation?

The equations should be:

$$s_y = (u\sin{\theta})t - \frac{1}{2}gt^2$$
and
$$s_x = (u\cos{\theta})t$$

where $s_y$ is the vertical displacement and $s_x$ is the horizontal displacement.

When the particle is at ground level, $s_y = 0$. Now solve for time t when that happens.

15. Aug 1, 2012

### Curious3141

OK, fast forwarding, and assuming you've found R to be given by: $R = \frac{u^2 \sin 2\theta}{g}$, put $\theta = 90^{\circ} - \alpha$. What is $\sin (180^{\circ} - \beta$) (where $\beta$ represents any angle)?