# Baseball Player hits Ball

1. Sep 14, 2016

### Veronica_Oles

1. The problem statement, all variables and given/known data
A baseball player throws a ball straight up and then catches ot 2.4 s later at the same heiggt from which he threw it. Determine the initial velocity and maximum height of ball?

2. Relevant equations

3. The attempt at a solution
Im confused about how to find inital velocity. I know its 2.4 s and the acceleration due to gravity is 9.8m/s^2 [d] however I feel like I am missing information. Is Final velocity 0?

2. Sep 14, 2016

If the ball went up and came down in 2.4 seconds, when is the ball at its highest point? Also how fast is the ball going at the top of the arc as it reverses direction?

3. Sep 14, 2016

### olivermsun

I believe you have all the information you need.

However, when thinking about the problem, try to think about a few key things:
1) How is the velocity of the ball related to the initial velocity and the acceleration due to gravity? (Do you have an equation for this?)
2) What's the velocity of the ball at the instant it achieves its maximum height?
3) Does it take more or less time for the ball to reach its max height, compared to the time it takes to fall back to the same height from which it was originally thrown?

4. Sep 14, 2016

### Staff: Mentor

What Relevant Equations do you know from your course materials? Which equations pertain to projectile motion?

5. Sep 17, 2016

### Veronica_Oles

Vi = Vf - (at)
Vi = 0 - (-9.8)(1.2)
Vi = 11.76m/s[up]
Vi = 12m/s[up]

Is this the correct way to do it?

6. Sep 17, 2016

### J Hann

Yes, or equivalently:
h = 0 = V0 * t - 1/2 * g t^2 which gives you the same result, where t here is the time to return to its original position.

7. Sep 17, 2016

@Veronica_Oles There are two ways to get the maximum height from this equation $h=v_0 t-(1/2)g t^2$. The simplest way is to plug in for $v_0$ and let t=1.2. (since you know the ball is at the top with velocity v=0 at $t=1.2$.) The more complicated way is to see that the graph of $h$ vs. $t$ is a parabola and has a maximum value since the coefficient of the $t^2$ term is negative. It is a good algebraic exercise to write the parabola in the form $h=A(t-t_0)^2+h_1$ by completing the square. $t=t_0=1.2$ is the axis of symmetry and $h_1$ turns out to be the maximum height. I would suggest simply writing $v_i$ as $v_i=(1.2)(9.8)$ if you work the extra algebra though, or you are going to see round-off errors and your axis of symmetry won't be exactly $t=1.2$.