Non-linear function with two unknown constants and one variable

[ETBunce]

Am am presented with the problem:

\(\displaystyle
h(t) = c - (d - 4t)^2
\)
At time t = 0, a ball was thrown upward from an initial height of 6 feet. Until the ball hit the ground, its height, in feet, after \(\displaystyle t\) seconds was given by the function \(\displaystyle h\) above, in which \(\displaystyle c\) and \(\displaystyle d\) are positive constants. If the ball reached its maximum height of 106 feet at time \(\displaystyle t = 2.5\), what was the height, in feet, of the ball at time \(\displaystyle t = 1\)?

The answer I am given is 70, but I don't know how to reach that answer. I've tried to solve for the constants but I keep hitting dead ends. I wish I could give more information other than the graph below but that's all I have. Help would be much appreciated, thanks.

I imagine the function if graphed would look something like this:

View attachment 2398

 

[lfdahl]

Hi, there

Thankyou for sharing your problem with us at the MHB.

$h(t) = c -(d-4t)^2$

Try to write down $h(0)$: You know from the initial condition, that $h(0) = 6$ (feet)
Then you should consider mathematically the expression at time $t = 2,5$ where the ball reaches its maximum height (i.e. find the derivative of $h(t)$)

 

[DavidCampen]

Actually, the problem is designed to simplify into solving a system of 2 linear equations in 2 unknowns.

 

[ETBunce]

$h(t) = c -(d-4t)^2$

Try to write down $h(0)$: You know from the initial condition, that $h(0) = 6$ (feet)
Then you should consider mathematically the expression at time $t = 2,5$ where the ball reaches its maximum height (i.e. find the derivative of $h(t)$)

Looks like the ball's height increases by 100 feet over the course of 2.5 seconds. Where do I go from there?

 

[lfdahl]

Looks like the ball's height increases by 100 feet over the course of 2.5 seconds. Where do I go from there?

You still need to find $c$ and $d$, the constants in the expression for $h(t)$.

Therefore, I have asked you to write down the two equations: $h(0) = ...$ and $h'(t=2.5) = 0$

Please give it a try (Wink)

 

[ETBunce]

You still need to find $c$ and $d$, the constants in the expression for $h(t)$.

Therefore, I have asked you to write down the two equations: $h(0) = ...$ and $h'(t=2.5) = 0$

Please give it a try (Wink)

Sorry, I'm confused. Derivatives are used in calculus, right? The problem I am presented with is from a SAT practice test. According to my understanding, I shouldn't need to use calculus tools to solve any of the problems in the SAT.
Should I go learn more about derivatives before trying to solve this problem?
I also have tried working out the equations, and I can't figure out how to solve for $c$ and $d$.

 

[lfdahl]

Sorry, I'm confused. Derivatives are used in calculus, right? The problem I am presented with is from a SAT practice test. According to my understanding, I shouldn't need to use calculus tools to solve any of the problems in the SAT.
Should I go learn more about derivatives before trying to solve this problem?
I also have tried working out the equations, and I can't figure out how to solve for $c$ and $d$.

I´m sorry, I am not familiar with the SAT practice test, but as far as I can see, you do need some knowledge on derivatives of a one-variable function. Anyway, what I was asking you about is the following:

(1). $h(0) = c -(d-4\cdot0)^2 = c-d^2 = 6$

(2). $h'(t) = (-4)\cdot2(d-4t)$ and use $h'(2.5) = 0$ so $d -4\cdot2.5 = 0$

From (2) you can determine $d$. Then you can insert the value for $d$ in (1) and find $c$.

Once you know $c$ and $d$, you can calculate $h(1) = ...$

 

[MarkFL]

If we write the height function in vertex form:

\(\displaystyle h(t)=-16\left(t-\frac{d}{4}\right)^2+c\)

and observe that we are given the vertex $(2.5,106)$

We then immediately have:

\(\displaystyle h(t)=-16\left(t-2.5\right)^2+106\)

And so what is $h(1)$?