# Ball motion

1. Dec 10, 2008

### Throwback24

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

Using the appropriate kinematic equations and initial values, determine the position and velocity at t=2.40 s.

(x, y)=0, 0
(vi, 0)=(25.0 m/s, 65.0 degree)

a.Horizontal Velocity of the initial velocity

b.Horizontal/Vertical Velocity at 2.40 s

c.Horizontal position at 2.40 s

2. Relevant equations

3. The attempt at a solution

a.(25 m/s)(Cos 65)

b.

c. X=(25 m/s)(2.40 seconds)

How do I get Horizontal/Vertical Velocity at 2.40 seconds?

2. Dec 10, 2008

### rigory

someone correct me if i am wrong, but i believe that you would want to break the horizontal/vertical velocity vector into into its components and then use either trig or pythag to find the avg velocity.

3. Dec 10, 2008

### Throwback24

But it's asking for the velocity @ 2.40 seconds. Don't you break up vectors into components?

4. Dec 10, 2008

### Gregg

For something moving 25ms-1 at an angle of 65 degrees to the horizontal you use. 25cos65 for the horizontal and 25sin65 for the vertical. use appropriate formula for the given times to find its velocity

5. Dec 10, 2008

### Throwback24

I'm wrong on C, it has to be:

25 m/s (Cos 65)(2.40 m)

I still don't get B. Gregg, isn't that how I solve for Initial velocity?

6. Dec 10, 2008

### Gregg

You can use
$$v = u + at$$ you know u, you know a and you are given t.

7. Dec 10, 2008

### Throwback24

I know V is Velocity, A is Acceleration and T is time. What is U? I've never seen it.

8. Dec 11, 2008

### Gregg

u is initial velocity, v0

9. Dec 11, 2008

Vf = Vi + at
or
V = V0 + at