Understanding Projectile Motion: Question on Angle and Distance

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The discussion centers on understanding the calculations of projectile motion when a ball is thrown at a 35-degree angle. The horizontal velocity is calculated using the cosine of 35 degrees, resulting in 20.5 m/s, while the vertical velocity uses the sine of 35 degrees, yielding 14.3 m/s. There is confusion about why the angle is referenced differently for horizontal and vertical components, with clarification that the vertical component can also be derived using the complementary angle of 55 degrees. The participants emphasize that both angles can be used interchangeably due to the geometric relationships in right triangles. The conversation concludes with a resolution of the initial confusion regarding the angle usage in projectile motion calculations.
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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?
 
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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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