# Range Calculations

1. Nov 25, 2016

### baba_944

1. The problem statement, all variables and given/known data
Here's a similar problem to the one in my book (I don't want to post the one in my book as I don't want to cheat):

The gravitational acceleration on Mars is 3.719[m/s^2].
The gravitational acceleration on Earth (excluding air friction and other places) is 9.81.

the initial velicty is this given in polar coordinates: 30<45o

How far will a ball go on Earth compared to on Mars? Use range calculations

2. Relevant equations

I want to say this is it:

R = vi2/g * sin2(theta)

3. The attempt at a solution

I did the following:

convert polar coordinates to rectangle coordinates:

x = 22cos(90)
y = 22sin(90)

x = 0
y = 22

R = (0,22)^2/9.81 * sin2(theta)

R = (0,22)^2/3.71 * sin2(theta)

That's all I got and that doesn't seem right.

Last edited: Nov 25, 2016
2. Nov 25, 2016

### Staff: Mentor

What direction is implied if the polar angle is 90°? What range can you expect for such a projectile?

Edit: Ah. I see that you've changed the angle and magnitude.

Last edited: Nov 25, 2016
3. Nov 25, 2016

### PeroK

Are you sure you mean polar coordinates? That doesn't make much sense.

Do you mean you are given an initial velocity and an initial angle?

4. Nov 25, 2016

### baba_944

I'm an idiot. Hold, up let me put the polar coordinates I have in my book to here. 90 degrees doesn't make senses as it's horizontal/vertical.

EDIT: Yea, I mean polar coordinates. Apologies for the polar coordinate format. If you want, I can take a screenshot of my book.

Now I'm unsure. It looks like polar coordinates to me.

5. Nov 25, 2016

### Staff: Mentor

Do whatever you need to do to show/describe the scenario.

6. Nov 25, 2016

### baba_944

I'm uploading a screenshot fro my book. Sorry I didn't do this before, I thought you guys could understand from my own example. Plus I want to solve it on my own after getting help.

7. Nov 25, 2016

### haruspex

I think baba means that if you view the launch in the vertical plane and express the initial velocity vector in polars then you get magnitude 30 (m/s?) and angle 45 degrees.

8. Nov 25, 2016

### baba_944

Here's the photo from the book:https://i.sli.mg/I2aK7f.jpg

I'm teaching myself physic so is it OK to ask you guys some questions every now and again?

9. Nov 25, 2016

### Staff: Mentor

Certainly. As long you follow the posting rules you can ask homework or homework-like questions here. If you just need to discuss something conceptually you might start a discussion in one of the technical forums.

Regarding the problem posted in this thread, you should make sure that you always attach units to any values that are specified or results that you present. A bare number is often meaningless without the associated units.

So we now understand that you are launching a projectile at an angle of 45° with a launch speed of 30 m/s, and doing so in two scenarios. One where the acceleration due to gravity is 9.83 m/s2 and one where it is 9.76 m/s2. What results does your range equation give you for those two scenarios?

Edit: Fixed launch speed. I thought I saw 40 m/s, now it seems to be 30 m/s. I await developments

10. Nov 25, 2016

### haruspex

Sure.
You only quoted an equation for range. That equation involves doubling the angle. You did not quote equations for converting from polar to Cartesian. Those do not double the angle. You can either work from first principles, using the correct initial horizontal and vertical velocities, or just apply the range equation. Since you are trying to teach yourself, I suggest you would find it most fruitful to work from first principles, with unknown angle and initial velocity, and derive the range equation for yourself.

Where did you get 22 from? In your question statement you mentioned 30. In the image from the book it is not legible, but it looks like either 10 or 40.

11. Nov 25, 2016

### PeroK

Your question about Mars is better. That one from your book is very silly. Huascaran is the highest mountain in Peru and is not very flat on top!

12. Nov 25, 2016

### baba_944

Ok, so just for clarification sakes:
dgs = degrees

30<45dgs

30 = initial velocity squared

45 = projectile's angle (plug in to theta)

NH: R= (30)^2[m/s]/9.76 *sin2(45)

NP: R = (30)^2[m/s]/9.83 * sin2(45)

sin2(theta) means two things I think

1: times by two or squared

2: Trigonometric identity: sin(theta) * cos(theta)

Going with the former:

NH: 130.41[m/s] (I rounded to the nearest tenth).

NP: 129.48[m/s]

So NH in Peru has the greater distance compared to the North Pole.

13. Nov 25, 2016

### Staff: Mentor

For a bit more clarity, the range equation is: $R = \frac{v^2}{g} sin(2θ)$. The "2" multiplies the angle, and is part of the sine function argument.

Note that if you click on the $\Sigma$ icon in the edit panel header, a menu of special symbols and Greek letters is made available. You can select special characters from that menu to use in your post. Items such as θ and the degree symbol ° are there.

14. Nov 25, 2016

### baba_944

Thank you. So is my calculations accurate?

15. Nov 25, 2016

### Staff: Mentor

No, you took the sine of 45° and then multiplied the result by 2. The angle itself needs to be doubled before you take the sine.