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Resolving max height with calculus

  • Thread starter gimpycpu
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  • #1
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Homework Statement


A ball is thrown at 32m/s vertically what is the max height the ball can attain.


Homework Equations


no equation is given except initial speed


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

Thank you for your help

Jonathan
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


A ball is thrown at 32m/s vertically what is the max height the ball can attain.


Homework Equations


no equation is given except initial speed


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

Thank you for your help

Jonathan
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:
  • #3
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What can't you use the equation for kinetic/potential energy?
 
  • #4
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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
LCKurtz
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Your answer is correct but you could certainly improve the writeup. I will insert suggested improvements in red.

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

f(0) = 32 so
##v(0)=32##

speed = f(x) = -9.8x + 32

##v(t) = -9.8t+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
##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
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Thank you for your help

I had no idea how to type correctly on the forum
 
  • #7
LCKurtz
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Thank you for your help

I had no idea how to type correctly on the forum
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
Ray Vickson
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Thank you for your help

I had no idea how to type correctly on the forum
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 [tex] s(t) = -\frac{1}{2} 9.8 t^2 - 32 t.[/tex] 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
What can't you use the equation for kinetic/potential energy?
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.
 
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