How Do You Calculate Initial Velocity and Maximum Height in Projectile Motion?

[Veronica_Oles]

Homework Statement

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?

Homework Equations

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?

 

[Charles Link]

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?

 

[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?

 

[gneill]

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

 

[Veronica_Oles]

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?

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?

 

[J Hann]

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?

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.

 

[Charles Link]

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?

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.

@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 $t=1.2$.