Velocity of the ball when it hits the ground, limits?

Homework Statement

[/B]
The path of a baseball relative to the ground can be modelled by the function
$d(t)=−t^2+8t+1$
, where d(t)represents the height of the ball in metres, and t represents time in seconds.
1. What is the speed of the ball when it hits the ground?

The Attempt at a Solution

$d(t)=-t^2+8t+1$

Using the quadratic to find the zeros, so I can find at what time the ball hits the ground.

$x=\frac {-b± \sqrt {b^2-4ac}} {2a}$

$x=\frac {-(8)± \sqrt {(8)^2-4(-1)(1)}} {2(-1)}$

x=-0.123 or x=8.123

Time can't be negative, therefore the ball htis the ground at 8.123 seconds.

I'm not sure how to find the velocity though. I thought I could take the first derivative of d(t)=−t2+8t+1 to find it but I couldn't figure it out that way. I think they want me to use limits somehow but I don't really know how.

Related Calculus and Beyond Homework Help News on Phys.org
How do you get the equation of velocity from the equation you were given? try not to change the t and x.

How do you get the equation of velocity from the equation you were given? try not to change the t and x.
I'm don't know. My teacher said to use the time that I got for when the ball hits the ground (x=8.123) and limits to solve for the velocity and I'm not sure how to do that.

I'm don't know. My teacher said to use the time that I got for when the ball hits the ground (x=8.123) and limits to solve for the velocity and I'm not sure how to do that.
Although the answer is correct, it is associated to the wrong variable. The function you are given is distance as a function of time. IOW, the independent variable here is t (time). Hence, the variable x in the quadratic equation above should be replace by t.

You were correct in taking the derivative of the given function since velocity is defined as what?

Think of what you have so far and what you can do with the clue.

Although the answer is correct, it is associated to the wrong variable. The function you are given is distance as a function of time. IOW, the independent variable here is t (time). Hence, the variable x in the quadratic equation above should be replace by t.

You were correct in taking the derivative of the given function since velocity is defined as what?

Think of what you have so far and what you can do with the clue.
Isn't velocity just the first derivative of a function with respect to time? I could take the derivative of the function and then subsitute in t= 8.123? This seems like an easier way to do it instead of using limits.
$d(t)=-t^2+8t+1$
$d'(t)+-2t+8$

$v(t)=d'(t)$
$v(t)=-2(8.123)+8$
$=-8.246$ m/s

If this is correct, is there another way to solve it using limit formula?

Isn't velocity just the first derivative of a function with respect to time? I could take the derivative of the function and then subsitute in t= 8.123? This seems like an easier way to do it instead of using limits.
$d(t)=-t^2+8t+1$
$d'(t)+-2t+8$

$v(t)=d'(t)$
$v(t)=-2(8.123)+8$
$=-8.246$ m/s

If this is correct, is there another way to solve it using limit formula?
Use the definition of a derivative. What is it?

This is a good place to understand the definition of a derivative.

Use the definition of a derivative. What is it?
$f'(x)=\frac {lim}{h→0} \frac {f(x+h)-f(x)} {h}$

Wasn't sure how to format the limit correctly with latex.

So i'd just use this and get the answer f'(x)=-2t+8, plug in my value for time when the ball hits the ground and then I have the velocity.

yup!. why is the velocity negative? This is also something to ponder.

yup!. why is the velocity negative? This is also something to ponder.
Would it have something to do with the direction of the ball?

Yes. That is the reason. Remember that gravity is acting on the ball. Hence, the ball is traveling in the townward direction when v is negative. One more thing. A common question that can be asked is the following. Find the maximum height the ball reaches. How would you do this?

Yes. That is the reason. Remember that gravity is acting on the ball. Hence, the ball is traveling in the townward direction when v is negative. One more thing. A common question that can be asked is the following. Find the maximum height the ball reaches. How would you do this?
I tthink I would take the derivative of d(t) and set it equal to 0 to find the maximum.

Yes. You find the time t when velocity is 0, then substitute this value of t into the position function. I assume you are taking Calculus 1? This is a very common question that can be asked on a quiz or even a test.

Yes. I assume you are taking Calculus 1? This is a very common question that can be asked on a quiz or even a test.
I guess it would be calculus 1. I'm taking this course online because I never took calc in highschool and now I want to go to university and the program I'm interested in requires a calc credit in grade 12. Thanks for the help, I don't have a teacher to go to as everything is done online so people like you help a lot.

NP. Are you using a book to learn from?

NP. Are you using a book to learn from?
I'm actually on the last unit and my exam is next week haha. I've been learning from the lessons the school provides online and youtube videos and some khan academy.

Although you may be getting some idea of calculus. Video lectures are not sufficient enough in regards to learning a subject. They are a supplement. I suggest you purchase a book or 2 on Calculus.

https://www.amazon.com/dp/B00GMPZBGA/?tag=pfamazon01-20

This is an older edition of Thomas Calculus and a really beautiful book. It is of the computation variety, unlike Spivak and Apostol, but it develops the ideas of calculus really well. MathWonk has written an extensive review on this book. Highly recommend, especially for self study.

https://www.amazon.com/dp/B000L3UO2A/?tag=pfamazon01-20

Calculus book by Moise. A first rate mathematician. It similar to Courant calculus, but a tad easier. Great balance between application and theory. This will prepare you to read an Analysis text such as Rudin.

I suggest you read and work from Thomas, and dip into Moise.

Although you may be getting some idea of calculus. Video lectures are not sufficient enough in regards to learning a subject. They are a supplement. I suggest you purchase a book or 2 on Calculus.

https://www.amazon.com/Calculus-Analytic-Geometry-George-Thomas/dp/B00GMPZBGA?crid=OYRR9TQK4C6J&keywords=calculus+and+analytic+geometry+thomas&qid=1547686943&sprefix=thomas+calculus+with+ana,aps,213&sr=8-3&ref=sr_1_3

This is an older edition of Thomas Calculus and a really beautiful book. It is of the computation variety, unlike Spivak and Apostol, but it develops the ideas of calculus really well. MathWonk has written an extensive review on this book. Highly recommend, especially for self study.

https://www.amazon.com/Calculus-complete-Edwin-Moise/dp/B000L3UO2A?crid=VM7GRK1IO4ON&keywords=moise+calculus&qid=1547687173&sprefix=moise+calcul,aps,264&sr=8-1-spell&ref=sr_1_1

Calculus book by Moise. A first rate mathematician. It similar to Courant calculus, but a tad easier. Great balance between application and theory. This will prepare you to read an Analysis text such as Rudin.

I suggest you read and work from Thomas, and dip into Moise.
Thank you!! I'll take a look at those!

Ray Vickson
Homework Helper
Dearly Missed
Isn't velocity just the first derivative of a function with respect to time? I could take the derivative of the function and then subsitute in t= 8.123? This seems like an easier way to do it instead of using limits.
$d(t)=-t^2+8t+1$
$d'(t)+-2t+8$

$v(t)=d'(t)$
$v(t)=-2(8.123)+8$
$=-8.246$ m/s

If this is correct, is there another way to solve it using limit formula?
You ARE using limits here! The derivative is defined as a particular kind of limit. Just look in any calculus textbook (or, if you do not have a textbook, look on-line under the key words "derivative" or "velocity").

Last edited: