Question

Parabola Question! Help! I just need to know how to solve.

Answer 1

A person standing on top of a 275 foot building throws a baseball vertically upward. The quadratic function is: s(t)=-16t^2+64t+275
t stands for seconds after it was thrown
How many seconds until it reaches maximum height?

Answer 2

@ganeshie8 ? :D

Answer 3

I'm concerned that there is also a note from my teacher saying it is necessary to show I used a certain equation, but i'm not sure which

Answer 4

@whpalmer4 or @robtobey ?

Answer 5

@Agent47

Answer 6

@TheSmartOne ?

Answer 7

I know how to find the vertex but i don't understand how time is coming into play!

Answer 8

what is the vertex?

Answer 9

i have only x is 2

Answer 10

y is found by k= f(h), right?

Answer 11

Vertex can be found by -b/2a

for the equation in the form of ax^2 + bx + c = 0

Answer 12

isn't that what was given?

Answer 13

the vertex is
\(\sf\Large (\frac{-b}{2a}, f(\frac{-b}{2a}))\)

Answer 14

ok so how do i use the vertex to find the time part? i'll do the vertex now...

Answer 15

but the variable is not x... what is the variable being used (in place of x) and what does it represent?

Answer 16

(2,335)? or I do that wrong?

Answer 17

2 is in place of x.. so 2 seconds?

Answer 18

I mean t^ lol

Answer 19

the answer will be the y-value of the vertex

Answer 20

oh

Answer 21

was my y value correct?

Answer 22

also, 335 is wrong.

2 is correct.

Plug t = 2 into your equation

Answer 23

oh...

Answer 24

ok one second

Answer 25

your y-value is close to the actual value, you must have made a small error when calculating it

so re-check your work :)

Answer 26

339?

Answer 27

correct! :)

Answer 28

yay!

Answer 29

so 339 seconds to the top?

Answer 30

thats an awful long time...

Answer 31

remember that is seconds, not minutes :P

Answer 32

i still couldn't throw hard enough for that! thank you,though!

Answer 33

Anytime! :D

Answer 34

just making sure- I shouldn't subtract the height of the building, right?

Answer 35

nope because we're dealing with how long it took to get it's heighest point :)

we're not dealing with how high it went up from where it was thrown :)

Answer 36

oh ok :)

Answer 37

@quixoticideals

Your parabola is \[h(t) = -16t^2+64t+275\]which more generally can be written as \[h(t) = -\frac{1}{2}gt^2+v_0 t + h_0\]where \(g\) is acceleration due to gravity (on the surface of the Earth, \(g\approx 32 \text{ ft/s}^2\)
\(v_0\) is the velocity with which the ball is launched in the vertical direction, a positive value indicating that it goes up
\(h_0\) is the initial height of the ball (here, the height of the building off of which the ball is thrown)
and \(h(t)\) is just notation telling us that we can find the height \(h\) at any time \(t\) we choose, by plugging our value of \(t\) into the formula.

As the squared term in the parabola equation has a negative coefficient, this will be a parabola which opens downward. We can find the \(x\)-coordinate of the vertex (which will be the apex of the ball's flight) with \[x=-\frac{b}{2a}\]if we have written our parabola in the form \[y = ax^2 + bx + c\]which we have done, just using different letters. Here our vertex \(x\)-coordinate would be \[t = -\frac{v_0}{2a} = -\frac{64}{2*(-16)} = 2\]
That means at \(t= 2\) (seconds) the ball is at its highest point in the flight.

How high is that point? Well, we plug \(t=2\) into our equation for the height:
\[h(2) = -16(2)^2+64(2)+275 = -64+128+275 = 339\]
or 339 feet off the ground (the reference point is the ground, NOT the top of the building, because we have that "+275" in our equation reflecting the height of the building).

For a bit of extra credit, when does the ball actually hit the ground? That will tell us the total time of flight. If we threw the ball from the ground, the parabola would have a \(y\) value (or \(h(t)\) value) of 0 at the instant we throw the ball, and at the instant the ball lands. With this scenario we have here, while our equation will still give two values of \(t\) that make \(h(t)=0\), notice that one of them is a negative value of \(t\), seemingly suggesting that the ball starts on the ground before it is thrown! While the height of the ball is modeled by a parabola, we only use the section of the parabola between \(t=0\) and the time at which the ball lands — the rest of it is not a valid model of the height of the ball.

So, we want to find a positive value of \(t\) that makes \[h(t) = 0 = -16t^2+64t+275\]
If we use the quadratic formula, we get values of \(t \approx - 2.603\) and \(t \approx 6.603\). We disregard the negative solution, as discussed, and now we know that the ball is in flight from \(t=0\) to \(t\approx 6.603\) or 6.603 seconds. I've attached a plot of the height of the ball (y-axis) vs. time (x-axis).

By the way, if you were standing on a 25+ story building, you would have no trouble throwing the ball high enough for it to stay in flight for several seconds! Just dropping it off the top, with no upward velocity, would result in the ball taking a bit more than 4 seconds to reach the ground. Clearly, if the ball is moving upward for the first part of the flight, it will take even longer, right?

Answer 38

Beautiful explanation, @whpalmer4 :-)