# Thread: Derivative function: ball thrown vertically has height s = 2 + 20t - 5t^2

1. ## Derivative function: ball thrown vertically has height s = 2 + 20t - 5t^2

The ball moved after the throw vertically in relation to the ground under the law
s = 2 + 20t - 5t^2 (s = distance from the ground)

a) From what height was the ball thrown?
b) What was the initial speed of the ball movement?
c) Was the ball thrown up or down?
d) How many seconds after the throw the ball was furthest away from the ground?
e) How much was the maximum distance of the ball from the ground?
f) How many seconds after the throw the ball landed on the ground?
g) What was the speed of the ball at the moment of falling?

I did the most basic thing -5*2t+20 = -10t+20 (s')
and from that I get -10 meter/sec^2 (s'') (Acceleration is -10 meter/sec^2 which means that the object was thrown up, is this answer for C and B?)

-10t+20=0 which gives me t=2sec (Ball stops at 2. sec?, answer for F?)
-10*0+20=20 meter/sec (20 meter/sec answer for G?)
But I don't understand how can I answer the other questions, what do I have to do?

2. There are basic principles you must understand. Quiz/Refresher.

s(t) = 2 + 20t - 5t^2 -- This describes the height of the ball.
s(0) = What?
s(m) = 0 means what?

v(t) = s'(t) = 20 - 10t -- This describes the velocity of the ball. As can be seen, it is linear and decreasing.
v(0) = What?
v(n) = 0 means what?

a(t) = -10 -- This describes the acceleration of the ball. As can be seen, it is constant.

You will need to get a better translation in order to differentiate between g and b.

3. Originally Posted by MagnusOir
The ball moved after the throw vertically in relation to the ground under the law

. . .s = 2 + 20t - 5t2

where s is the distance of the ball from the ground

a) From what height was the ball thrown?
b) What was the initial speed of the ball movement?
c) Was the ball thrown up or down?
d) How many seconds after the throw the ball was furthest away from the ground?
e) How much was the maximum distance of the ball from the ground?
f) How many seconds after the throw the ball landed on the ground?
g) What was the speed of the ball at the moment of falling?

I did the most basic thing -5*2t+20 = -10t+20 (s')
What is this? Do you mean that you took the derivative, so "-5*2t+20=-10t+20" means the following:

. . . . .$s'(t)\, =\, 20\, -\, 5\,(2t)\, =\, 20\, -\, 10t$

Originally Posted by MagnusOir
and from that I get -10 meter/sec^2 (s'') (Acceleration is -10 meter/sec^2 which means that the object was thrown up, is this answer for C and B?)
How did you obtain this value? How is negative acceleration (due to gravity?) provide the direction of the throw? How would a negative rate of change in the height indicate an upward trajectory?

Originally Posted by MagnusOir
-10t+20=0 which gives me t=2sec (Ball stops at 2. sec?, answer for F?)
-10*0+20=20 meter/sec (20 meter/sec answer for G?)
But I don't understand how can I answer the other questions, what do I have to do?
It might help to work in an orderly fashion, thinking as we go, and using what we've learned back in algebra.

(i) You are given a "height" function, s(t) = 2 + 20t - 5t^2, where the height s is given in terms of the time t. At what time t is the ball thrown? Evaluating the height function at this time t, what value do you get for the height s? So from what height was the ball thrown?

(ii) You know that the derivative of a "position" function (such as the given "height" function) is the velocity function (or the "speed"). At the time determined in the previous paragraph (i), what then was the value of the velocity?

(iii) Is this value for the velocity, from (ii), positive or negative? What does this indicate regarding the direction of the throw?

(iv) You were given a "height" function which is a parabola. You know, from algebra, where the max/min value of a parabola resides. What is this max/min value? Is it a max or a min? What then is the distance? At what time did this distance occur?

(v) You know the value for "height away from ground" when the ball is on the ground. Setting the height equal to this value and solving (just like when you did this exercise back in algebra), what time do you obtain?

(vi) I'm not sure what "at the moment of falling" means in part (g), but I will guess that it intends to say something along the lines of "at the moment the ball touched the ground". So evaluate the "velocity" function (that is, the derivative) at the time you obtained in (v). What velocity to you get?

If you get stuck, please reply showing all you work in following the above step-by-step instructions. Thank you!

4. Originally Posted by stapel
(i) You are given a "height" function, s(t) = 2 + 20t - 5t^2, where the height s is given in terms of the time t. At what time t is the ball thrown? Evaluating the height function at this time t, what value do you get for the height s? So from what height was the ball thrown?
So S=2+20*0-5*0^2=2 From height 2

Originally Posted by stapel
(ii) You know that the derivative of a "position" function (such as the given "height" function) is the velocity function (or the "speed"). At the time determined in the previous paragraph (i), what then was the value of the velocity?
S= -10*0+20=20m/s

Originally Posted by stapel
(iii) Is this value for the velocity, from (ii), positive or negative? What does this indicate regarding the direction of the throw?
Up

Originally Posted by stapel
(iv) You were given a "height" function which is a parabola. You know, from algebra, where the max/min value of a parabola resides. What is this max/min value? Is it a max or a min? What then is the distance? At what time did this distance occur?
-10t+20=0
t=2

Originally Posted by stapel
(v) You know the value for "height away from ground" when the ball is on the ground. Setting the height equal to this value and solving (just like when you did this exercise back in algebra), what time do you obtain?

Originally Posted by stapel
(vi) I'm not sure what "at the moment of falling" means in part (g), but I will guess that it intends to say something along the lines of "at the moment the ball touched the ground". So evaluate the "velocity" function (that is, the derivative) at the time you obtained in (v). What velocity to you get?
S= -10*0+20=20m/s

Are these right or even close?

5. Originally Posted by stapel
(i) You are given a "height" function, s(t) = 2 + 20t - 5t^2, where the height s is given in terms of the time t. At what time t is the ball thrown? Evaluating the height function at this time t, what value do you get for the height s? So from what height was the ball thrown?
Originally Posted by MagnusOir
So S=2+20*0-5*0^2=2 From height 2
I will guess from the above string that you mean the following:

At the time the ball is thrown, the time is t = 0. So I'll evaluate the height function at t = 0:

. . . . .s(0) = 2 + 20(0) - 5(0)^2 = 2 + 0 - 0 = 2

Then the ball was thrown from a height of s = 2 [units].
Note: Because you have not posted the entire exercise, we cannot know what [units] are correct. You'll need to make the necessary change in your hand-in work. Also, "s" and "S" are two completely different variables, so you'll need to make that correction before handing in your work, too.

Originally Posted by stapel
(ii) You know that the derivative of a "position" function (such as the given "height" function) is the velocity function (or the "speed"). At the time determined in the previous paragraph (i), what then was the value of the velocity?
Originally Posted by MagnusOir
S= -10*0+20=20m/s
I'm not sure what is going on here...? The original function, s(t), is not at all the same thing as the derivative function, s'(t). But you seem to think you are using the original function, which you here call "S"...?

Originally Posted by stapel
(iii) Is this value for the velocity, from (ii), positive or negative? What does this indicate regarding the direction of the throw?
Originally Posted by MagnusOir
Up
Yes, the initial direction of motion was upwards.

Originally Posted by stapel
(iv) You were given a "height" function which is a parabola. You know, from algebra, where the max/min value of a parabola resides. What is this max/min value? Is it a max or a min? What then is the distance? At what time did this distance occur?
Originally Posted by MagnusOir
-10t+20=0
t=2
My guess is that you are referencing the vertex of the parabola, which is the local max for this function. To find the location of the local max, you set the derivative, s'(t) = 20 - 10t, equal to zero, and solved. If correct, your answer almost certainly still needs units, however.

Originally Posted by stapel
(v) You know the value for "height away from ground" when the ball is on the ground. Setting the height equal to this value and solving (just like when you did this exercise back in algebra), what time do you obtain?
Originally Posted by MagnusOir
I'm sorry, but I don't know what this means...?

What is the height above the ground when the ball is on the ground?

Originally Posted by stapel
(vi) I'm not sure what "at the moment of falling" means in part (g), but I will guess that it intends to say something along the lines of "at the moment the ball touched the ground". So evaluate the "velocity" function (that is, the derivative) at the time you obtained in (v). What velocity to you get?
Originally Posted by MagnusOir
S= -10*0+20=20m/s
I'm sorry, but I have no idea what you're doing here...?

Originally Posted by MagnusOir
Are these right or even close?
Without knowing what was your reasoning, which function you're using at a given point, at what value you're evaluating that function or why, it's kinda hard to say. Sorry!

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