# Help with Projectiles

1. Sep 15, 2007

### Heat

1. The problem statement, all variables and given/known data

A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 degree above the horizontal. The shot hits the ground 2.08s later. You can ignore air resistance.

What is the x-component of the shot's velocity at the end of its trajectory?
What is the y-component of the shot's velocity at the end of its trajectory?
2. Relevant equations

x = v initial cos theta time

3. The attempt at a solution

I managed to get the x and y component of acceleration.

but for this I would suppose its going to be 12cos51 2.08 = 15.71

I am lost...as this is wrong.

Last edited: Sep 15, 2007
2. Sep 15, 2007

### learningphysics

The 2.08 should be s, not m/s right?

What is the initial x and y component?

What is the acceleration of the shot put?

3. Sep 15, 2007

### Heat

x-component of the shot's velocity at the beginning of its trajectory

7.55m/s

y-component of the shot's velocity at the beginning of its trajectory
9.32

y-component of the shot's acceleration while in flight
-9.80
x-component of the shot's acceleration while in flight
0

from what I got thusfar.

now I need to find the x and y component of the shot's velocity at the end of its trajectory

Last edited: Sep 15, 2007
4. Sep 15, 2007

### learningphysics

cool. you have everything you need... you have initial velocity... acceleration... time... you should be able to get final velocity.

EDIT: why do you have -9.32 instead of 9.32?

5. Sep 15, 2007

### Heat

so the equation would be vx = v initial +gt

vx = 11.99 + (9.8)(2.08)

= 32.37

but now how do I get x and y, (I don't have degree for the final velocity)

6. Sep 15, 2007

### learningphysics

No... vertical acceleration affects vertical velocity... horizontal acceleration affects horizontal velocity...

7. Sep 15, 2007

### Heat

sorry , but i don't see where you are going with this...

8. Sep 15, 2007

### learningphysics

The x-component of acceleration is 0... so does the x-component of velocity change?

9. Sep 15, 2007

### Heat

no...same for y also....

10. Sep 15, 2007

### learningphysics

Ah... but there is a y-component to acceleration... calculate the final y-velocity using initial y velocity and the y-acceleration.

11. Sep 15, 2007

### learningphysics

vx = v0x + ax*t

vy = v0y + ay*t

12. Sep 15, 2007

### Heat

vx = v0x + ax*t

vy = v0y + ay*t

vx = 7.55 + (0)(2.08)

vy = 9.32 + (-9.8)(2.08)

vx = 7.55

vy = -11.064

Which is correct, thank you learning physics.

Last edited: Sep 15, 2007
13. Sep 15, 2007

### Heat

also,

How far did she throw the shot horizontally?

this is what I did

I decided to use the following equation: x-x initial = t/2 (vinitial + vx)

I know the vf and vi for x and y.

so I decided to do a^2 + b^2 = C^2

so the velocity for initial would be 11.99
for final it would be 13.39

so now its

x-0 = 2.08 /2 (11.99+13.39)

x= 26.39

so that would be distance right? but I am wrong :(

14. Sep 15, 2007

### learningphysics

you're making it too complicated... horizontal displacement, you only need horizontal velocity... you know the horizontal velocity, you have the time... you know acceleration is 0...

15. Sep 15, 2007

### Heat

I tend to do that.

x-0 = 7.55(2.08) + .5........

final being 15.70 :)