# Homework Help: Ball in gravity: how high does the ball go? Initial velocity?

1. Apr 12, 2004

### mayo2kett

i know this is a really simple problem but i'm kinda stuck on it since i don't know what to do with the equation i'm given ...

Having won the game, a baseball pitcher hurls the ball up in the air. Its height in feet t seconds after the throw is given by
h(t)=5.9+69t-16t^2

How high does the ball go?
What is its initial velocity?

2. Apr 13, 2004

Are you familiar with derivatives?

All you have to do is maximize that function and evaluate the derivative of that function at 0.

3. Apr 13, 2004

### holly

Thowing the ball up: h=initial height + v0*t - (1/2)gt^2. That's the usual gravity accel equation-thing.

69 is the initial velocity. It must be feet per second, because 16=1/2 of 32 and g=32 ft/s^2, or so they say. 5.9 is the initial height, they gave it to you.

The total time from when it goes up and comes back to the initial height is when height=initial height,
so,
height-initial height = 0 = v0 * t - 1/2 gt^2, solve for t, the time.

When you get that, 1/2 of that is on the way up to its max height.
So put 1/2 of that time into the first equation to get the max height.

4. Apr 13, 2004

### turin

Holly's method allows you to do it without derivatives. That is, you just compare it to the standard form of the "kinematical equation."

5. Apr 13, 2004

### mayo2kett

i have learned derivatives but i was trying to learn how using the formula...

i now understand how to get the initial velocity and height ( i didn't know they gave then to me like that), but i'm a little stuck on the max height... this is the work i did:
0 = v0 * t - 1/2 gt^2
0 = 69t-(1/2)(9.81)t^2
t=14.067
(1/2)14.067=7.03

then:
h=hi+ v0*t - (1/2)gt^2
h=5.9+69(7.03)-(1/2)(9.81)(7.03)^2
=248.56
but this answer seems to be wrong... i'm not sure what i did, was it just a multiplying error?

6. Apr 13, 2004

### turin

You have to identify the numbers in the kinematic equation, which is strictly mathematical (pre-physics, if you will). You are imposing unwarratned physics. Perhaps it would help you to imagine that you are on some alien planet. Then, consider the kinematical equation in the form that holly suggested. (Of course, also as holly pointed out, this is representative of earth in the units given, but it is a good exercise to consider the more fundamental meaning of the kinematical equation.) In other words, 9.81 is inappropriate in this case.

7. Apr 13, 2004

### mayo2kett

ok so i understand why i shouldn't included gravity in this but how else can i solve for t?? i think i'm making this much harder than it should be, but i never learned anything about kinematics...

8. Apr 13, 2004

### mayo2kett

whouldn't it just be easier if i took the derivative of this: h(t)=5.9+69t-16t^2
so: h(t)'=69-32t
but what do i do from here to get the max height? our physics teacher has only taught us using formulas...

9. Apr 13, 2004

### Staff: Mentor

What you did wrong was to put your own value for g in the equation. The original equation was perfectly fine. Why did you put g = 9.81?

As Holly explained, this equation uses feet for distance units, not meters. So g = 32 ft/s^2.

To use Holly's method, do this. Find the times where the height is the original height. Why? That equation is trivial to solve! You were on the right track, but you modified the given equation for some reason.

So: 0 = 69t-16t^2 = t (69-16t)
The solutions are t = 0 and t = 69/16 = 4.31 sec.

Now you take it from here. Using the original equation for height, of course!

10. Apr 13, 2004

### Staff: Mentor

The maximum height is where the derivative (the velocity!) is zero. Find the value of "t" that gives zero velocity. Then find the height at that time.

11. Apr 13, 2004

### mayo2kett

you guys are awsome thanks so much... i understand what i was doing wrong now, i just didn't understand holly's explanation correctly... we've never learned that gravity could be anything besides 9.81. :)

12. Apr 13, 2004

### holly

Make sure you aren't including me in your "thanks," because I didn't learn any of "my" method by myself. Turin, marcus, Doc Al, cookiemonster and the other Brains of the forum showed me how to do it before. It's these Brains I am mighty thankful for...when one of them answers you, you can rest easy...THEY ARE GREAT!!!

13. Apr 14, 2004

### turin

I think you earn the primary credit for this one, holly. It is an arrogant person who assumes self sufficiency for their own knowledge.

14. Apr 14, 2004