# Another cannonball question

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1. Mar 13, 2015

### cptstubing

1. The problem statement, all variables and given/known data
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

2. Relevant equations
initial velocity = 50 m/s

3. 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?

2. Mar 13, 2015

### SammyS

Staff Emeritus
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.

3. Mar 13, 2015

### Merlin3189

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

4. Mar 17, 2015

### cptstubing

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!

5. Mar 17, 2015

### SteamKing

Staff Emeritus
You'll get over it.

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