Resolving max height with calculus

The "correct" answer to the OP's question is that the ball reaches its maximum height when its velocity is 0, and that happens t seconds after it is thrown, where t is the solution to -9.8t+32=0, i.e., t=32/9.8 seconds. Substitute that value of t into the position function to get s=23.60 meters.)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
  • #1
gimpycpu
6
0

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
 
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  • #2
gimpycpu said:

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

gimpycpu said:
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
Thank you for your help

I had no idea how to type correctly on the forum
 
  • #7
gimpycpu said:
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
gimpycpu said:
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
voko said:
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.
 
Last edited:

Related to Resolving max height with calculus

1. What is the purpose of using calculus to resolve maximum height?

The purpose of using calculus to resolve maximum height is to find the exact value of the maximum height of an object or projectile. Calculus allows us to find the maximum value of a function, in this case, the height function, by finding the derivative and setting it equal to zero.

2. What are the key concepts in calculus that are used to resolve maximum height?

The key concepts used in resolving maximum height with calculus are derivatives, optimization, and critical points. Derivatives help us find the rate of change of the height function, optimization helps us find the maximum value of the function, and critical points are the points where the derivative is equal to zero.

3. How is calculus used to determine the maximum height of a projectile?

Calculus is used to determine the maximum height of a projectile by finding the derivative of the height function and setting it equal to zero. This will give us the critical point, which is the point where the object reaches its maximum height. We can then plug this value into the original function to find the maximum height.

4. Are there any limitations to using calculus to resolve maximum height?

Yes, there are limitations to using calculus to resolve maximum height. Calculus assumes that the height function is continuous and differentiable, which may not always be the case in real-life scenarios. Additionally, calculus may not be applicable for objects that are affected by external forces such as air resistance.

5. How does resolving maximum height with calculus relate to real-world applications?

Resolving maximum height with calculus is applicable in many real-world applications, particularly in physics and engineering. It allows us to accurately predict the maximum height of objects or projectiles, which is crucial in designing structures such as bridges or determining the trajectory of a rocket. It also has practical uses in fields such as sports, where knowing the maximum height of a jump or throw can be helpful.

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