Calculating Velocity of a Cannonball at a Given Height

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Homework Help Overview

The problem involves calculating the velocity of a cannonball shot upward from the ground, given its height function s(t) = 50t - 4.9t^2. The specific question is about finding the velocity when the cannonball reaches a height of 100 meters on its ascent.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to set the height equation equal to 100 to find the time t. There is a focus on solving the resulting quadratic equation rather than relying on trial and error. Questions about the proper method to determine velocity at that height are raised.

Discussion Status

Some participants acknowledge the validity of using trial and error while others emphasize the importance of finding an analytical solution. There is recognition of the quadratic nature of the equation, and some guidance is offered regarding the approach to solving it.

Contextual Notes

Participants express frustration about the categorization of the thread, indicating a potential misunderstanding of the problem's nature. There is also mention of personal reflection on problem-solving approaches after a break.

cptstubing
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Homework Statement


A Cannonball is shot upward from the ground into the air at t=0 sec. With initial velocity of 50m/s. Its height above the ground in metres is given by s(t)=50t-4.9t^2 .
----What is the velocity of the cannonball when it is 100m above the ground on the way up?
"says the book" Hint: set s(t) = 100

Homework Equations


initial velocity = 50 m/s

The Attempt at a Solution


I don't want the answer, I just want to know how it is worked out, because I know how to do it by trial and error.
If I set s(t)=100 like the book says, then 100=50t-4.9t^2
I can't solve this equation without knowing t, am I right?
Anyhow, by trial and error, i figured out t=2.73, and I think velocity at 100m = 36.5m/s...
But what is the proper way to do this without using trial and error like I did?

ie.
The question I should ask myself first is what?
1. What is t at 100m?
2. Velocity = ? at 100m?
 
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cptstubing said:

Homework Statement


A Cannonball is shot upward from the ground into the air at t=0 sec. With initial velocity of 50m/s. Its height above the ground in metres is given by s(t)=50t-4.9t^2 .
----What is the velocity of the cannonball when it is 100m above the ground on the way up?
"says the book" Hint: set s(t) = 100

Homework Equations


initial velocity = 50 m/s

The Attempt at a Solution


I don't want the answer, I just want to know how it is worked out, because I know how to do it by trial and error.
If I set s(t)=100 like the book says, then 100=50t-4.9t^2
I can't solve this equation without knowing t, am I right?
Wrong. In fact, to find the time, t, at which the cannon ball is 100m above the ground, you need to solve this.

It's a quadratic equation.
Anyhow, by trial and error, i figured out t=2.73, and I think velocity at 100m = 36.5m/s...
But what is the proper way to do this without using trial and error like I did?

ie.
The question I should ask myself first is what?
1. What is t at 100m?
2. Velocity = ? at 100m?
 
I think you deserve credit for solving by trial and improvement. I often do this myself when I can't see a simple analytic solution.

As Sammy says, it is a quadratic equation, so it's just a maths issue to solve it.

WikiP explains it here
 
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Thanks to all for the help.
I actually figured this out very quickly after a relaxing weekend.
Fresh eyes matter.
On a side note, whoever moved this thread from the 'Pre-calculus Math' section and into 'Introductory Physics Homework' section, really bruised my ego!
 
cptstubing said:
On a side note, whoever moved this thread from the 'Pre-calculus Math' section and into 'Introductory Physics Homework' section, really bruised my ego!

You'll get over it. :wink:
 

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