How to make a proof for complementary angles of elevation.

[Shanyn]

Homework Statement

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.

Homework Equations

sin θ = cos 90 - θ

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.

 

[Curious3141]

Homework Statement

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.

Homework Equations

sin θ = cos 90 - θ

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.

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.

 

[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.

 

[Curious3141]

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.

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

 

[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.?

 

[azizlwl]

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.?

Recheck your equations.
Vertical displacement is zero for the range.

 

[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?

 

[voko]

Why the minus sign?

 

[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 :)

 

[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.

 

[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.? :)

 

[voko]

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

 

[azizlwl]

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?

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

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

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

 

[Curious3141]

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.?

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.

 

[Curious3141]

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 :)

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)?