1. ## Quadratic Functions

Hey guys,

Maybe you could just check this for me and let me know if its correct, and help where i require it

Ok so a ball is projected vertically from ground - modelled by:

$h=-5t^2 + 10t$

Interval, $t = 0s$ to $t=2.5s$ - plot this on your calc. setting window for the time interval.

1, Find height of ball after 0.5s = 3.75m

2, Find the 2 times when the ball is @ h = 0 - explain method = 2secs & ?? Help

3, What is the maximum value of t which this
$f$ is a model for height of ball = explain

4, Find the vertex of the Parabola and explain method = using

$y=ax^2 + bx + c$ (1) and $( -b\frac{b}{2a}, C - \frac{b^2}{4a} )$ (2)

Model
$h= -5t^2 +10t$ means using (1)

$a= -5, b= 10 and c= 0$

Vertex co-ordinates are:

$\frac{-b}{2a} = \frac{10}{2 x -5} = \frac{10}{-10} = -1$ (For x Co-ord)

$C - \frac{b^2}{4a} = \frac{10^2}{4 x -5} = -5$ (For y Co-ord)

5, What is the velocity of the ball @ the vertex = Zero velocity

6, Equation of Velocity-Time is given by
$V = 10 - 10t$ find the velocity of the ball @ $t=0.5s$ & $t=1.5s$ = 1m/s and -1m/s

7, Explain what the plus and minus signs mean in Q6, what they indicate? = direction of travel

8, How do you find the speed of the ball frm velocity? = the gradient, here it is
$1/(1.5-2) = -2m/s^-1$ which is a speed of 2ms^1 in a downward direction

9, By inserting h=1.8 , explain why solutions of quadratic:

$t^2 - 2t + 0.36 = 0$

gives the times at which the ball is at a height of h = 1.8 above the ground?

Thanks for all your help in advance

2. Originally Posted by ADY
...a ball is projected vertically from ground - modelled by: $h=-5t^2 + 10t$

1, Find height of ball after 0.5s = 3.75m
If you mean that h(0.5) = 3.75, then I agree with your answer.

Originally Posted by ADY
2, Find the 2 times when the ball is @ h = 0 - explain method = 2secs & ??
Plug "zero" in for "h", and apply the Quadratic Formula.

Originally Posted by ADY
3, What is the maximum value of t which this $f$is a model for height of ball = explain
Since the ball is assumed to hit the ground and stop (rather than continuing downward through the Earth), what is the last t-value for which the h-values make physical sense? (You'll use your answer to part (2) for this.)

Originally Posted by ADY
4, Find the vertex of the Parabola and explain method = using $y=ax^2 + bx + c$ (1) and $( -b\frac{b}{2a}, C - \frac{b^2}{4a} )$(2)

Model $h= -5t^2 +10t$ means using (1)

$a= -5, b= 10 and c= 0$

Vertex co-ordinates are:

$\frac{-b}{2a} = \frac{10}{2 x -5} = \frac{10}{-10} = -1$ (For x Co-ord)
$C - \frac{b^2}{4a} = \frac{10^2}{4 x -5} = -5$ (For y Co-ord)
I'm not sure what they mean by their first "method", unless they're wanting you to complete the square to find the vertex...?

Using the formula is often easier, and you might want to do "method 2" first, so you can be sure of your answer to "method 1". You might want to check your work on this, by the way: -(10) does not equal "10".

Originally Posted by ADY
5, What is the velocity of the ball @ the vertex = Zero velocity
Yes, the velocity at the peak of its path is zero; it has stopped going up, and is about to come back down.

Originally Posted by ADY
6, Equation of Velocity-Time is given by $V = 10 - 10t$ find the velocity of the ball @ $t=0.5s$ & $t=1.5s$ = 1m/s and -1m/s
How did you arrive at 10 - 10(1/2) equalling 1? Or 10 - 10(3/2) equalling -1?

Originally Posted by ADY
7, Explain what the plus and minus signs mean in Q6, what they indicate? = direction of travel
A positive velocity indicates forward (in this case, upward) motion; a negative velocity indicates backward (in this case, downward) motion.

Originally Posted by ADY
8, How do you find the speed of the ball frm velocity? = the gradient, here it is $1/(1.5-2) = -2m/s^-1$ which is a speed of 2ms^1 in a downward direction
The speed is the velocity, but without regard to direction. You'd take the absolute value, not the derivative.

Originally Posted by ADY
9, By inserting h=1.8 , explain why solutions of quadratic: $t^2 - 2t + 0.36 = 0$ gives the times at which the ball is at a height of h = 1.8 above the ground?
Just follow the instructions: Plug "1.8" in for "h", which means "the times t when the height h is 1.8 meters", and rearrange (by putting everything together on one side of the "equals" sign and dividing through by the appropriate value) to get the equation into the form they showed.

3. Originally Posted by stapel
If you mean that h(0.5) = 3.75, then I agree with your answer.
No i mean 0.5 seconds. "Find the height of the ball after 0.5s" - Don't you make this 3.75 then?

Originally Posted by stapel
Plug "zero" in for "h", and apply the Quadratic Formula.
Cant i just find the answer of the graph? In which case when h=0, the two times are 2secs and ?

Originally Posted by stapel
Since the ball is assumed to hit the ground and stop (rather than continuing downward through the Earth), what is the last t-value for which the h-values make physical sense? (You'll use your answer to part (2) for this.)
How do i know where the ground is though?

Originally Posted by stapel
I'm not sure what they mean by their first "method", unless they're wanting you to complete the square to find the vertex...?

Using the formula is often easier, and you might want to do "method 2" first, so you can be sure of your answer to "method 1". You might want to check your work on this, by the way: -(10) does not equal "10".
So are my co-ordinates wrong? - If so what do you make them, and could you please explain your method?

Originally Posted by stapel
Yes, the velocity at the peak of its path is zero; it has stopped going up, and is about to come back down.
Woo!

Originally Posted by stapel
How did you arrive at 10 - 10(1/2) equalling 1? Or 10 - 10(3/2) equalling -1?
Ok, i have no idea what i did, so we have $v = 10 -10t$ - "Find the velocity of the ball @ t = 0.5s and t = 1.5s

Originally Posted by stapel
A positive velocity indicates forward (in this case, upward) motion; a negative velocity indicates backward (in this case, downward) motion.

Originally Posted by stapel
The speed is the velocity, but without regard to direction. You'd take the absolute value, not the derivative.
Is there a formula i could show this working?

Originally Posted by stapel
Just follow the instructions: Plug "1.8" in for "h", which means "the times t when the height h is 1.8 meters", and rearrange (by putting everything together on one side of the "equals" sign and dividing through by the appropriate value) to get the equation into the form they showed.

Ok so it looks like $t^2 - 2t + 0.36 = 1.8$ where do i go from here?

Thanks for your help

4. Originally Posted by ADY
Originally Posted by stapel
If you mean that h(0.5) = 3.75, then I agree with your answer.
No i mean 0.5 seconds. "Find the height of the ball after 0.5s" - Don't you make this 3.75 then? Mr F says: Stapel has agreed with you! She worded it like this because the way you expressed your answer was poor.

Originally Posted by stapel
Plug "zero" in for "h", and apply the Quadratic Formula.
Cant i just find the answer of the graph? In which case when h=0, the two times are 2secs and ? Mr F says: You've been told how to do it ....

Originally Posted by stapel
Since the ball is assumed to hit the ground and stop (rather than continuing downward through the Earth), what is the last t-value for which the h-values make physical sense? (You'll use your answer to part (2) for this.)
How do i know where the ground is though? Mr F says: It should be clear that the ground corresponds to h = 0.

Originally Posted by stapel
I'm not sure what they mean by their first "method", unless they're wanting you to complete the square to find the vertex...?

Using the formula is often easier, and you might want to do "method 2" first, so you can be sure of your answer to "method 1". You might want to check your work on this, by the way: -(10) does not equal "10".
So are my co-ordinates wrong? - If so what do you make them, and could you please explain your method? Mr F says: Did you check your work like was was suggested?

Originally Posted by stapel
Yes, the velocity at the peak of its path is zero; it has stopped going up, and is about to come back down.
Woo!

Originally Posted by stapel
How did you arrive at 10 - 10(1/2) equalling 1? Or 10 - 10(3/2) equalling -1?
Ok, i have no idea what i did, so we have $v = 10 -10t$ - "Find the velocity of the ball @ t = 0.5s and t = 1.5s Mr F says: Stapel is saying, kindly, that you have botched the basic arithmetic. Go back and do it right.

Originally Posted by stapel
A positive velocity indicates forward (in this case, upward) motion; a negative velocity indicates backward (in this case, downward) motion.

Originally Posted by stapel
The speed is the velocity, but without regard to direction. You'd take the absolute value, not the derivative.
Is there a formula i could show this working? Mr F says: Do what stapel said to do.

Originally Posted by stapel
Just follow the instructions: Plug "1.8" in for "h", which means "the times t when the height h is 1.8 meters", and rearrange (by putting everything together on one side of the "equals" sign and dividing through by the appropriate value) to get the equation into the form they showed.

Ok so it looks like $t^2 - 2t + 0.36 = 1.8$ where do i go from here? Mr F says: No! Where did you get this from? You certainly didn't follow the instruction very carefully.

Substitute h = 1.8 into h = -5t^2 + 10t:

-5t^2 + 10t = 1.8 => 5t^2 - 10t + 1.8 = 0.

Now divide both sides by 5.

Thanks for your help
..

5. Originally Posted by mr fantastic;
Mr F says: No! Where did you get this from? You certainly didn't follow the instruction very carefully.

Substitute h = 1.8 into h = -5t^2 + 10t:

-5t^2 + 10t = 1.8 => 5t^2 - 10t + 1.8 = 0.

Now divide both sides by 5...
You used the wrong equation, that's why i decided to start that other thread because this is not very well laid out,.

By inserting h=1.8 , explain why solutions of quadratic:

gives the times at which the ball is at a height of h = 1.8 above the ground?

6. Originally Posted by ADY
You used the wrong equation, that's why i decided to start that other thread because this is not very well laid out,.

By inserting h=1.8 , explain why solutions of quadratic:

gives the times at which the ball is at a height of h = 1.8 above the ground?
No, I have used the correct equation.

You have misundertsood the question. If you did the last small thing I left for you to do (divide through by 5) you would realise this.

7. I have and now i feel silly!

Thank you for your help and guidence Mr.F!

This one is solved. Apart from i dont know how to find the speed of the ball from the velocity?

Thank you!