How high is the stone at its highest? Relationships are given

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The discussion focuses on determining the maximum height of a rock based on its motion equations, x=25t and y=20t-5t^2. The participants confirm that the trajectory of y is a quadratic curve, and the peak height occurs at the midpoint of the time values where y equals zero. One participant suggests using the vertex formula or completing the square to find the maximum height, while another mentions that finding the midpoint of the time values is sufficient. The conversation emphasizes that while taking the derivative is a valid method, it may not be necessary for this particular problem. Ultimately, the participants agree on the approach to find the stone's highest point.
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Homework Statement



A rock moves so that its coordinates at the time t given by the relationships

x=25t
y=20t-5t^2

How high is the stone at its highest?

Homework Equations



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The Attempt at a Solution



I got the distance the stone flies (horizental ground) by setting y=0, that gave x=100m
If that is correct, how should I proceed?
 
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What curve is y with regard to t?
 
voko said:
What curve is y with regard to t?

A quadratic curve?
 
Yes it is. You know the two points where it meets the t axis. What can be said about the location of its peak?
 
voko said:
Yes it is. You know the two points where it meets the t axis. What can be said about the location of its peak?

It's above and between the two points.
 
So as you know it's exactly midway, you know the value of t to plug into the equation and get the height.
 
voko said:
So as you know it's exactly midway, you know the value of t to plug into the equation and get the height.

So I take the value of t I got from finding out the flying distance and divide it by 2?

If that is correct I got my answer! Thank you.
 
Well, you could do it the "proper" way by taking the derivative, equating it to zero, etc.

But since you already know the roots of the equation, you can use the fact that the apex of the parabola is always right in the middle.
 
I would not consider taking the derivative to be the "proper" way for a problem like this. Rather, complete the square, so you get something like y= h- (x- a)^2. That is h when x= a, h minus something otherwise. That is, y= h is the maximum value of y.
 
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