Understanding Projectile Motion: Question on Angle and Distance

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Homework Help Overview

The discussion revolves around understanding projectile motion, specifically the relationship between the angle of projection and the horizontal and vertical components of velocity. The original poster questions the reasoning behind using different angles for horizontal and vertical components when a ball is thrown at a 35-degree angle upward.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the geometric interpretation of angles in projectile motion, questioning why the vertical component is associated with 55 degrees while the horizontal component is associated with 35 degrees. Some participants attempt to clarify the use of trigonometric functions in this context.

Discussion Status

The discussion is ongoing, with participants providing visual aids and attempting to clarify the relationship between the angles and the components of velocity. There is a mix of understanding and confusion regarding the application of trigonometric principles in this scenario.

Contextual Notes

Participants express uncertainty about the angles used in calculations and the underlying assumptions about the geometry of projectile motion. There is a mention of the original poster being new to the concepts involved.

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Homework Statement


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?
 
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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.
 
Kruum said:
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.

Ok I do not get how the angel with the horizontal is 35 while the vertical is 55 because you aimed 35 degree upward.
 
I hastely put together a picture of the situation.
http://www.aijaa.com/img/b/00677/3628783.jpg
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:
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
 
psmarz said:
C = 25m/s
B = 20.5m/s
A = 14.3

So did you find out, why you can also use 35 degrees instead of 55? Can you show me how you got those figures?
 
(35 sin) 25 = 14.3

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

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

Yep!
 
Thank you for helping me figure it out.
 
  • #10
psmarz said:
Thank you for helping me figure it out.

You're very welcome!
 

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