# Resolving max height with calculus

1. Aug 19, 2012

### gimpycpu

1. The problem statement, all variables and given/known data
A ball is thrown at 32m/s vertically what is the max height the ball can attain.

2. Relevant equations
no equation is given except initial speed

3. The attempt at a solution
so my guess is since the gravity is 9.8m/s^2

if I integrate 9.8m/s^2 = 32 it gives me 2.1396
so after 2.1396 the ball should have its max height.

So I end up with this equation
height = 32*2.1396 - 9.8*2.1396^2
height = 23.60 meters

But I am soo uncertain about my answer, I have to use calculus to resolve it

Jonathan

2. Aug 19, 2012

### LCKurtz

Start out with the equation for acceleration: $a=-9.8$. Integrate that with respect to time twice. What do you get for velocity $v$ and position $s$? Don't forget to think about the constants of integration.

Last edited: Aug 19, 2012
3. Aug 19, 2012

### voko

What can't you use the equation for kinetic/potential energy?

4. Aug 19, 2012

### gimpycpu

no I can't use the kinect equation, but i found the answer with calculus

f(0) = 32 so
speed = f(x) = -9.8x + 32
so if I integrate this from 0 to 32/9.8 it gives me 52.24 meters

(-9.8t^2)/2 + 32x
(-9.8(32/9.8)^2)/2 + 32(32/9.8) = 52.24 meters

which is a mere aproximate

5. Aug 19, 2012

### LCKurtz

Your answer is correct but you could certainly improve the writeup. I will insert suggested improvements in red.

$s(t)=\frac {-9.8t^2}{2}+32t$
$s(\frac{32}{9.8})=52.24$

Even with those improvements, you have hand-waved evaluating the constants and evaluating the $t$ value for max height, which you seem to understand.

6. Aug 19, 2012

### gimpycpu

I had no idea how to type correctly on the forum

7. Aug 19, 2012

### LCKurtz

Of course, proper notation is more than typing. For example mixing up $x$ and $t$ in your equations.

Also, voko's suggestion for using KE/PE equations gives an easy solution. At the bottom the kinetic energy is $\frac 1 2 mv^2$ and at the top there is no KE but the potential energy is $mgh$. Setting those equal and solving for $h$ gives the same answer.

8. Aug 19, 2012

### Ray Vickson

He was not referring to the use of LaTeX, etc., but, rather, to the method of presentation. There would be nothing wrong in writing s(t) = -(1/2)(9.8)t^2 - 32t, or (1/2)*(9.8)*t^2 - 32*t for example, instead of $$s(t) = -\frac{1}{2} 9.8 t^2 - 32 t.$$ As long as you use brackets to make things clear, you should be OK---but do not "overuse" them either.

Generally, in Physics especially, it is a bad idea to use "x" or "y" to refer to time (even though time may be an unknown in your problem); usually we employ the letters "x" or "y" to refer to coordinates on a line, or in a plane or in space. Of course, using "x" as a time variable is not illegal, but it is unwise; certainly, you should use the same notation from start to finish of a problem (which you did not: you wrote (-9.8t^2)/2 + 32x, where x and t are supposed to be the same thing!) Writing up your work to make it easier to mark is always recommended.

RGV

9. Aug 20, 2012

### noclueatol

Perhaps because he's never heard of it. The fact that he's posting in the calculus section suggests that the problem is from a math book, not a physics book. And even in most freshman physics books, KE/PE concepts are introduced several chapters later than the kind of problem in the OP.

Last edited: Aug 20, 2012