When to Use Sin and Cos in Projectile Motion Calculations?

  • Thread starter Thread starter jensson
  • Start date Start date
  • Tags Tags
    Cos Sin
AI Thread Summary
In projectile motion calculations, sine and cosine are used to resolve the initial velocity into horizontal and vertical components based on the angle of launch. For a ball thrown at an angle of 20 degrees below the horizontal, the horizontal component is calculated using cosine (8 cos 20), while the vertical component uses sine (8 sin 20). This distinction arises from the definitions of sine and cosine in relation to the right triangle formed by the velocity vector, where sine corresponds to the opposite side (vertical) and cosine to the adjacent side (horizontal). Understanding this vector resolution is crucial for solving projectile motion problems accurately. Properly applying these principles allows for effective calculations of distance and height in projectile motion scenarios.
jensson
Messages
25
Reaction score
0

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)
 
Physics news on Phys.org
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.
 
Back
Top