1. ## quadratic formula/completing the square?

First off I this is for a review for my class due tomorrow. I lost my notes and I am completely lost. I don't know if this is algebra or not, there used to be a help-board I see that is gone which is a shame (or maybe I'm blind?).

A rock is thrown into the air from a bridge and falls to the water below. The height of the ball, h metres, relative to the water t seconds after being thrown is given by:

h(t)=-5t2+10t+35

a) Determine the maximum height of the rock above the water.
b) How long does it take the rock to reach the maximum height?
c) After how many seconds does the rock hit the water?

Thanks in advance, I really hate exam time, so stressful. :P

2. Are you sure that's all the information that the question includes? Usually they will give the height of the bridge or something similar, because both c) and a) are impossible to solve without knowing the height of the initial throw above the water.

Also, what level of math is this for? If you have taken intro calculus, I believe differentiation would really help here. We can find b) right now by taking the derivative of the function and setting it equal to 0 to find the maximum.

$
h(t)=-5t^{2}+10t+35
$

$
h'(t)=-10t+10=0
$

$
\frac{-10t}{-10}=\frac{-10}{-10}
$

$
t=1
$

i.e. It takes the rock 1 second to reach maximum height.

3. Originally Posted by risteek
First off I this is for a review for my class due tomorrow. I lost my notes and I am completely lost. I don't know if this is algebra or not, there used to be a help-board I see that is gone which is a shame (or maybe I'm blind?).

A rock is thrown into the air from a bridge and falls to the water below. The height of the ball, h metres, relative to the water t seconds after being thrown is given by:

h(t)=-5t2+10t+35

a) Determine the maximum height of the rock above the water.
b) How long does it take the rock to reach the maximum height?
c) After how many seconds does the rock hit the water?

Thanks in advance, I really hate exam time, so stressful. :P
This is deffinitely calc1.

Because no instructor would give you this question in algebra.

To find the maximimum hieght, of the thing, find the max of the function. You know how to find the maxes of functions right. Jus t set f'(x)=0.
So
$h'(x)=-10t+10$

set it to zero

$-10t+10=0$

$10t=10$
$t=1$
the thing is at a max at t=1

Plug 1 into the original position function to know how high it is gonna go, then add that height to the hieght of the bridge and let that sum be equal to h(t), solve for t, and you'll have how much time it took to reach the grounnd.

One more thing. This problam could have been solved with plain old algebra for the most part. If the guy wasn't on a bridge, you could just solve h(t) for 0 and then the time that the hieght would have been attained would simply be the average. You could have done the whole thing without calc, but I don't want to get long winded.

4. No that is all my teacher gave me. I am in a Grade 11. University/College Functions and Applications class.

Thank you for your help, though

5. Originally Posted by VonNemo19

This is deffinitely calc1.

Because no instructor would give you this question in algebra.

To find the maximimum hieght, of the thing, find the max of the function. You know how to find the maxes of functions right. Jus t set f'(x)=0.
Not true. This can be done in an algebra class. The maximum/minimum value of a parabola is at the vertex. Since the leading coefficient is negative, it will be a maximum. To find the maximum height put it in the vertex form $y = a(x - h)^2 + k$ by completing the square:

\begin{aligned}
h(t) &= -5t^2 + 10t + 35 \\
h(t) &= -5(t^2 - 2t) + 35 \\
h(t) &= -5(t^2 - 2t + 1) + 35 + 5\\
h(t) &= -5(t - 1)^2 + 40\\
\end{aligned}

The vertex in $y = a(x - h)^2 + k$ would be (h, k), so the maximum height is 40m at t = 1 sec.

01

6. Originally Posted by risteek
c) After how many seconds does the rock hit the water?
Find the zeros of the function:

\begin{aligned}
-5t^2 + 10t + 35 &= 0 \\
-5(t - 1)^2 + 40 &= 0 \\
-5(t - 1)^2 &= -40 \\
(t - 1)^2 &= 8 \\
t - 1 &= \pm \sqrt{8} \\
t &= 1 \pm 2\sqrt{2} \\
t &= 1 + 2\sqrt{2} \approx 3.83 sec. \\
t &= 1 - 2\sqrt{2} \approx -1.83 sec.
\end{aligned}

Reject the second solution because it's negative. It'll take about 3.83 seconds before the rock hits the water.

01

7. Thank you for all of your help. As I read through your posts my memory of the subject came back, thanks!