Projectile motion of a ball launch

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Homework Help Overview

The problem involves the projectile motion of a ball launched over level ground, described by a vertical equation of motion. Participants are tasked with finding the time of maximum height and maximum height, as well as initial and final vertical velocities.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the method of completing the square to find the maximum height and time, with one participant expressing uncertainty about their completion of the square. Others emphasize the importance of differentiating the equation to find stationary points where the gradient is zero.

Discussion Status

The discussion is ongoing, with participants clarifying the need to differentiate the equation before determining the time of maximum height. Some participants seek further explanation of the differentiation process and its significance in identifying maximum and minimum points.

Contextual Notes

There is a request for guidance on completing the square correctly, and participants are exploring the implications of differentiating the motion equation in the context of the problem.

southernbelle
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Homework Statement


On earth, a ball is launched over level ground. It's vertical equation of motion is
y(t) = -4.9(t2 - 8t - 20)

A) Find tymax and ymax
B) Find vyinitial and vyfinal


Homework Equations


y(t) = Yi + via + 1/2at2
Vf = vi + at


The Attempt at a Solution


I tried to complete this problem by completing the square. I'm not sure I completed the square correctly.

For the completed square I got:
-4.9(t-4)2 + 58.4
making tymax = 4s
and ymax = 58.4 m

To get tymax I set the equation equal to zero and factored
I got
0 = -4.9(t + 2) (t -10)
making ti = -2 s and
tf = 10 s

For Vyi I took the derivative of the y(t) equation and plugged in ti
Vyi = -9.8(-2) + 39.2
Vyi = 58.6 m/s

For Vyf I used the derivative and plugged in tf
Vyf = -9.8(10) + 39.2
Vyf = -58.8 m/s

If I did not complete the square in the first part, please show me how to do it rather than telling me a different way to go about the problem. Thank you!
 
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You must differentiate first and then set the result equal to zero. The gradient is obviously zero at maxima and minima.
 
I must differentiate before getting the t value?
Because I did differentiate before I got the Vi and Vf values.
 
Like I say, to find a maximum or minimum you need to find where the gradient is zero, therefore you must differentiate.
 
Can you explain that a little further? Because I have no idea what you're talking about.
 
The derivative of a function tells us something about how that function behaves at certain points. At stationary points like maximums and minimums the derivative is zero. Therefore to find the point where this function is maximum you must take its derivative, set it equal to zero and solve.
 
southernbelle said:
Can you explain that a little further? Because I have no idea what you're talking about.

You have to find the derivative of the function and set it to equall to zero.when you set it to equall to zero your final solution will be two numbers so if the graph goes from negative to positive you will have minimum and from positive to negative you will have maximum point.but If you graph it, it will be easy.

And one more derivative is a slope.
 

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