How to determine X and Y coordinates of launched ball?

[TexasCow]

Homework Statement

Lets say one were to launch a ball at 60 degrees at 80m/s. How would one determind the x and y coordinates at time intervals of the ball?

Homework Equations

I'm not sure of equations that should be used.

The Attempt at a Solution

 

[HallsofIvy]

1) F= ma

2) Force and acceleration are vectors: you can look at X (horizontal) and Y (vertical) separately.

Neglecting air resistance, the only force on the ball after it has been "launched" is gravity, -mg, vertically.

From this point, I can think of several different ways to procede but, since you have shown no work at all, I don't know which would be appropriate! If you have not taken calculus, then there are probably specific equations given in your textbook.

 

[TexasCow]

I'd like to use the equation:

Δx = Vavg x Δt
Δy = Voy Δt + ½ g Δt^2


The thing is, when I use these equations the values just keep getting bigger and bigger...I don't know what/how to manipulate the equation to get a negative parabola.

 

[GoldPheonix]

Homework Statement

Lets say one were to launch a ball at 60 degrees at 80m/s. How would one determind the x and y coordinates at time intervals of the ball?

Homework Equations

I'm not sure of equations that should be used.

The Attempt at a Solution

Relevant equations:

x = x_i + v_i t + \frac{1}{2} a t^2
y = y_i + v_i t + \frac{1}{2} a t^2

*NOTE: the velocity and acceleration must be the vector portions:

\Delta s = total displacement
s = \sqrt{x^2 + y^2}
x = s cos \theta
y = s sin \theta

However, velocity is a vector, like s, your displacement. So, you apply trig, record the correct velocity in the y-direction and in the x-direction.

 

[TexasCow]

Thanks Pheonix that helps a bunch. I'm having the same problem as before. I'm sure it's a simple mis understanding, but since gravity is the sole effect on the ball, the velocity obviously has to decrease towards the peak. Using the equations above, as time increases, so does velocity. How do I have the velocity values go back down as if I were to actually do the lab?

 

[Edgardo]

I'd like to use the equation:

Δx = Vavg x Δt
Δy = Voy Δt + ½ g Δt^2


The thing is, when I use these equations the values just keep getting bigger and bigger...I don't know what/how to manipulate the equation to get a negative parabola.

You are on the right track. But I suggest you first consider the following problem: You throw a ball vertically up in the air with the initial velocity Voy.
How does the equation look like that describes the motion of the ball?

 

[TexasCow]

I don't know...new to this whole Physics thing.

 

[Edgardo]

Use the equation you gave for Δy.
Suppose you throw a ball vertically up in the air with 80 m/s.
How does Δy look like?

 

[Dick]

I'd like to use the equation:

Δx = Vavg x Δt
Δy = Voy Δt + ½ g Δt^2


The thing is, when I use these equations the values just keep getting bigger and bigger...I don't know what/how to manipulate the equation to get a negative parabola.

Are you putting g=-9.8m/sec^2? Emphasis on the NEGATIVE.

 

[beadmaster]

agreeing with goldpheonix, the ket is to seterate the initial velocities into components with trig. & depending on question resolving the final velocity at the given time peroid with pythagoras

havn't done projectiles in a while though.

 

[turdferguson]

The velocity and acceleration must oppose each other as dick said