Question

Arianna kicks a soccer ball off the ground and into the air with an initial velocity of 42 feet per second. Assume the starting height of the ball is 0 feet. Approximately what maximum height does the soccer ball reach?

Answer 1

@ganeshie8
@whpalmer4
@zzr0ck3r

Answer 2

@UnkleRhaukus
@RadEn
@shamil98

Answer 3

@eliassaab @timo86m @chmvijay

Answer 4

\[h(t) = -\frac{1}{2}gt^2 + v_0t + h_0\]\[g=32\frac{\text{ft}}{\text{s}^2}\]\[v_0=42\frac{\text{ ft}}{\text{ s}}\]\[h_0 = 0\text{ ft}\]

The formula is that of a parabola, \[y = ax^2 + bx + c\]where the vertex is located at \(x = -\dfrac{b}{2a}\) Plug that value into the formula to find the corresponding y value for the vertex (except here we'll be calling it \(h\)).

Answer 5

what does that mean?

Answer 6

It means the function relating the height of the ball with respect to time is a parabola, and the maximum height will be at the vertex of the parabola.

Answer 7

yes is it 128 ft?

Answer 8

@whpalmer4

Answer 9

I'm afraid you are not even close...care to show me your work?

Answer 10

oops what the heck i don't know why i wrote 128 i meant 272 ft

Answer 11

@whpalmer4

Answer 12

You're getting even less close :-(

How about you show me your work?

Answer 13

hmmm. maybe my formula is wrong

Answer 14

what formula should i use?

Answer 15

I gave you everything you need...

Answer 16

it would probably be easiest if you just humored me and told me how you got these answers

Answer 17

i guessed

Answer 18

jk im not sure what I'm doing wrong then

Answer 19

is x=−b/2a the formula?

Answer 20

@whpalmer4

Answer 21

Okay, here's the height formula I gave you:

\[h(t) = -\frac{1}{2}gt^2 + v_0t + h_0\]When you plug in the various numbers specified in the problem, what is that formula?

Answer 22

OH okay thank you i got 224 ft

Answer 23

@whpalmer4

Answer 24

???

Answer 25

no, please just tell me what that formula looks like after you have inserted the values of \(g\), \(v_0\), and \(h_0\)

Answer 26

are you sure its not 224? i double checked and all.

Answer 27

@whpalmer4

Answer 28

I'm 100% positive. Look, either you show me your work, or I leave.

Answer 29

Once again:

height of the ball is given by \[h(t) = -\frac{1}{2}gt^2+v_0 t + h_0\]\(h_0\) is the initial height of the ball
\(v_0\) is the initial velocity of the ball
\(g\) is the acceleration due to gravity, \(32 \text{ ft}/\text{s}^2\)

Please write out the formula after you have inserted those values.

Answer 30

0(t)=−12(32)2+(42)t+(0)

Answer 31

thats what i get when i plug in the height velocity and acceleration

Answer 32

not -12, -1/2

and h(t) doesn't turn into 0(t)

Answer 33

\[h(t) = -\frac{1}{2}(32)t^2+42t+0\]\[h(t) = -16t^2+42t\]

Answer 34

so i have to solve for t?

Answer 35

@whpalmer4

Answer 36

No. That's a parabola. You're looking for the maximum height the ball reaches. This parabola (because the coefficient of the \(t^2\) term is negative) looks like an inverted bowl, with the top of the arc being at the vertex. The vertex of such a parabola can be very easily found by dividing the coefficient of the \(t\) term by twice the coefficient of the \(t^2\) term and changing the sign. That gives you the value of the independent variable (here, \(t\)) at the vertex. You then plug that value into the formula for the parabola to find the value of the dependent variable (here, \(h(t)\)), which will be the maximum height the ball reaches.

Answer 37

i got 27.5625

Answer 38

@whpalmer4

Answer 39

That is correct.

Answer 40

WOOOHOOO!

Answer 41

Thanks!

Answer 42

Here's a graph with the axes labeled and two lines showing t and h(t) at the vertex

Answer 43

Another way you could have found the answer in this particular case is to remember that a parabola is symmetrical about the perpendicular line connecting the directrix and the vertex. Because the initial height was 0, the two solutions where the height is 0 are located symmetrically about the vertex, and if you factored the equation, you would get \[-2t(8t-21)=0\]leading to \[t=0,t=\frac{21}{8}\]as the solutions where the height is 0. Halfway between them is \(t = \dfrac{21}{16}\) which is the same value you get from the \(-\dfrac{b}{2a}\) formula. However, if we had \(h_0\ne0\) then this approach would not work so conveniently (one of the solutions would involve a negative value of \(t\) rather than 0).

Answer 44

okay thank you so much for all your help @whpalmer4