# Question on Projectile

1. Feb 18, 2009

### psmarz

1. The problem statement, all variables and given/known data
The ball is now thrown with the same speed at an angel of 35 deg upward. How far does it travel?

My question is...

They have

vhor = (25 m/s) cos 35 deg = 20.5 m/s

and

The angel with the vertical is (90 - 35) = 55 deg
so...

vvert = (25 m/s) cos 55 deg = 14.3

It seems backwards to me for some reason because it says that the ball was thrown at a angel of 35 deg upward.

Can someone help me understand why 35 deg is tacked onto the horizontal while 55 is tacked onto the vertical?

2. Feb 18, 2009

### Kruum

Usually when talking about projectile motion the angel is to the horizontal line. So you aim 35 degrees up from the horizon. And if you take a closer look at the geometry of a triangle you'll see it is possible to use the same angel for horizontal and vertical speed. Just keep in mind that you can move vectors around.

3. Feb 18, 2009

### psmarz

Ok I do not get how the angel with the horizontal is 35 while the vertical is 55 because you aimed 35 degree upward.

4. Feb 18, 2009

### Kruum

I hastely put together a picture of the situation.
http://www.aijaa.com/img/b/00677/3628783.jpg [Broken]
a is the vertical speed, b is horizontal speed and c is a and b combined. Those two pictures mean the exact same thing, I've just moved a. How would you solve a from the picture on the right, if you only knew the value of c and the angel?

Last edited by a moderator: May 4, 2017
5. Feb 18, 2009

### psmarz

Thank you for the visual. I am new to the whole SOH, CAH, TOA and that visual helps greatly...

C = 25m/s
B = 20.5m/s
A = 14.3

6. Feb 18, 2009

### Kruum

So did you find out, why you can also use 35 degrees instead of 55? Can you show me how you got those figures?

7. Feb 18, 2009

### psmarz

(35 sin) 25 = 14.3

They did it by subtracting 35 from 90 and doing (55 cos)25 = 14.3

8. Feb 18, 2009

### Kruum

Yep!

9. Feb 18, 2009

### psmarz

Thank ya for helping me figure it out.

10. Feb 18, 2009

### Kruum

You're very welcome!