When to Use Sin and Cos in Projectile Motion Calculations?

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Homework Help Overview

The discussion revolves around understanding the application of sine and cosine in projectile motion calculations, specifically in the context of a ball being thrown from a building at an angle. The original poster seeks clarity on why these trigonometric functions are used for the horizontal and vertical components of motion.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the geometric interpretation of resolving vectors into components, discussing the relationship between the angle of projection and the use of sine and cosine. Questions about how to visualize this with triangles and the definitions of sine and cosine are raised.

Discussion Status

Some participants have provided insights into the geometric reasoning behind the use of sine and cosine, while others express uncertainty about vector resolution and drawing triangles. There is an ongoing exploration of how these concepts apply to the problem at hand.

Contextual Notes

The original poster indicates a lack of familiarity with vector resolution and geometric representation, which may affect their understanding of the discussion. Additionally, the context includes a specific problem involving projectile motion with given initial conditions.

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Homework Statement


This is a more general question... but here is a specific example in which I don't know why sin is used where it is and why cos is used where it is. I don't need the problem worked out or anything (I have it worked out by my professor), I just need to understand why sin and cos are used when they are.


A ball is tossed from an upper story window of a building. The ball is given an initial velocity of 8 m/s at an angle of 20 degrees below the horizontal. It strike the ground 2 seconds later. How far horizontally from the base of the building does the ball strike the ground? Find the height from which the ball was thrown. How long does it take the ball to reach a point 10 m below the level of launching?




Homework Equations



initial velcoity = 8 sin 20

y = initial y - 8 sin 20 (t) - 1/2 g t^2

v(y direction) = -8 sin 20 - gt

x = 8 cos 20 (t)

v(x direction) = 8 cos 20 (t)
 
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Pure geometry. Draw a triangle representing the initial velocity and you'll see why. The magnitude of the initial velocity is the hypotenuse of the triangle and the x and y components are the legs.
 
Initially, the velocity is at at angle of 20°.

So the initial vertical velocity is 8sin20 and the initial horizontal velocity is 8cos20.

It comes down to resolving vectors. Do you know how to do this?
 
initial velocity = 8 at 20 degrees below horizontal so
horizontal component Vx = 8*cos(20)
vertical component Vy = 8*sin(20) downward

Vertical part: y = initial y - 8 sin 20 (t) - 1/2 g t^2
v(y direction) = -8 sin 20 - gt

Horizontal part:
x = 8 cos 20 (t)

v(x direction) = 8 cos 20 [the (t) that was here is a mistake]
 
No, I don't know how to resolve vectors.. I'm taking calc I and physics concurrently so I'm a little behind. I also am not sure how to draw a triangle representing the initial velocity (or anything for that matter).
 
trig.jpg
 
Nice!

Okay, so I use cos in the x direction because the adjacent line to the angle known is in the x direction, and sin in the y direction because the opposite is in the y direction? This changes depending on the triangle and the angle known, yes?
 
Yes. At first you should use the definitions like cos A = adjacent/hypotenuse. After some practice you will be able to go straight to Vx = 8*cos(20).
 
Thank you so much!
 
  • #10
Make sure you are using angles relative to the horizontal axis. Otherwise the roles reverse.
 

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