# Projectile Motion ball height

1. Sep 22, 2009

### Lemniscate

1. The problem statement, all variables and given/known data
A small dense ball with mass 1.5 kg is thrown with initial velocity <5,8,0> m/s at time t= 0 from the origin (<0,0,0> m) Air resistance is negligible.

When the ball reaches its maximum height, what is its velocity?

2. Relevant equations
Momentum Principle
Average Velocity

3. The attempt at a solution
Though I have an idea of how to approach some of the questions regarding this problem, I have not been able to understand the following:
How do I know what the maximum height the ball reached is?

2. Sep 22, 2009

### Furby

What do you know about an object that reaches its maximum height?

Think about what occurs to the motion of the ball at and around that time frame.

3. Sep 23, 2009

### Lemniscate

It's speed is 0, so the velocity (vector) would be <0,0,0> m/s at its max ht...?

4. Sep 23, 2009

### raknath

dude the maximum height will be reached when the object has reached the distance which can be covered by throwing it at that angle, why should this be so confusing

The velocity vector will not be 0 at maximum height it would be the same, only direction of resultant will change based upon the energy that a body spends opposed to the acceleration.

Last edited: Sep 23, 2009
5. Sep 23, 2009

### Furby

That would be correct if the ball was thrown straight up, in a 1-D situation, without an x component to the velocity.

According to your question the ball was thrown with an initial velocity of <5,8,0> m/s. If at the maximum height the ball is not accelerating vertically, then (as we concluded) the vy is 0m/s.

Now reobserve the maximum height of the ball. After reading this, what are you left with?

6. Sep 24, 2009

### Lemniscate

The x-component of the velocity at its max ht is the same as the initial and the y-component of this vector is 0 since the ball is not acceleration vertically. Yes?

7. Sep 24, 2009

### Furby

I have to apologize on my previous wording, the ball is accelerating vertically, but the velocity is just 0. If it wasn't accelerating, then in it would just float there in the air. That is the point where the velocity that is positive, and thus rising in position, changes to negative, and thus begins to fall back towards the ground. So you're somewhat correct.

Correct. There is no acceleration in the horizontal direction, thus the x component of the velocity remains constant. That is the only velocity vector, and thus is your velocity.

8. Sep 24, 2009

### Lemniscate

So, just to make sure, my velocity is <5,0,0> m/s at this point?

Thank you for your help, by the way. :)

9. Sep 24, 2009

Yup! =)