# Thread: [SOLVED] Polynomial time/height

1. ## [SOLVED] Polynomial time/height

Hey guys, I need help with this!

A ball is dropped from top of a building. Its height from the ground (in feet) after t
seconds is given by H (t) = – 16 t^2 + 320.
a) Graph this function.
b) What is an appropriate domain for this function?
c) How tall is the building?
d) How high is the ball after traveling 2 seconds?
e) When does the ball reach the ground?
f) What is the range of this function?
g) Another ball is thrown upward from top of the same building. Give an example of
a function G where G(t) gives the height of the ball form the ground after t
seconds.

2. Originally Posted by OVechkin8
Hey guys, I need help with this!

A ball is dropped from top of a building. Its height from the ground (in feet) after t
seconds is given by H (t) = – 16 t^2 + 320.
a) Graph this function.
What techniques have you learned for graphing? There are many different ways to do this.

b) What is an appropriate domain for this function?
"Domain" refers to t values. Phyiscally, what are the possible times here?

c) How tall is the building?
Where was the ball when t= 0?

d) How high is the ball after traveling 2 seconds?
What is x(2)?

e) When does the ball reach the ground?
The ground is x= 0. Solve $-16t^2+ 320= 0$.

f) What is the range of this function?
What are the possible heights for the ball?

g) Another ball is thrown upward from top of the same building. Give an example of
a function G where G(t) gives the height of the ball form the ground after t
seconds.
The ball in this example was "dropped" which means initial speed 0. If it is "thrown upward" what can you say about the initial speed?

I would feel better about these, and might be able to give more help, if you showed some sign of having tried these your self. Do you understand that most of these problems require only integer arithmetic?

3. For the answers I got:

a. An inverted parabola in the form of -16(t^2-20)
b. the domain would be all (-oo, +oo)
c. the height is 320?
d. -16(2)^2 + 320 = 256
e sqrt(20)
f. range is [320, -oo)

I still need help with g.

4. Originally Posted by OVechkin8
For the answers I got:

a. An inverted parabola in the form of -16(t^2-20)
The graph is part of that parabola. See my response to (b)

b. the domain would be all (-oo, +oo)
No, it would not. You did not drop the ball until t= 0 and the ball will no longer be dropping after it hit the ground. Your graph should cover only that interval of time.

c. the height is 320?
Yes, very good.

d. -16(2)^2 + 320 = 256
Yes, very good.

e. sqrt(20)
Yes. $t= \sqrt{20}= \sqrt{(4)(5)}= 2\sqrt{5}$. And this is where the parabola stops.

f. range is [320, -oo)
No, the ball does NOT go through the earth!
Remember that this is not just a problem about the mathematical formula $y= -16x^2+ 320$. It is a problem about a balling falling.

I still need help with g.
Motion problems, with constant acceleration g= -16 always give a function like $h= -16t^2+ V_0 t+ H_0$ where $H_0$ is the height at time t= 0 and $V_0$ is the velocity at t= 0. In your original problem, $V_0$ was 0 because the ball was "dropped". If the ball is "thrown up" its velocity at t= 0 is some positive number.

5. AHH, so the range would be [320, 0]?

For the domain are you talking about positive values, so from 0, to where it touches the x axis?

6. Originally Posted by OVechkin8
AHH, so the range would be [320, 0]?
Well, I would put it the other way around- [0, 320]. The range is simply as set of numbers. The fact that the ball goes from 320 to 0 feet doesn't change the fact that the range is the interval from 0 to 320.

For the domain are you talking about positive values, so from 0, to where it touches the x axis?
Yes. Which you have already calculated as being $[0, 2\sqrt{5}]$.

7. Awesome, thank you!

I'm still confused about g, would G(t) = -16t^2 + xt + h?? or something to that effect?