Question

PLz answer question asap includes formula

Answer 1

do u understand question

Answer 2

i need help asap

Answer 3

does anyone know how to do it at lest

Answer 4

if it gives you all the info just try plugging them into the formula.
Now if you will excuse me i will be back after i'm done eating.
when either you will have solved this question or someone else has helped you.

Answer 5

The question includes a formula, but that won't solve the problem directly. Try this:
\[v _{f}^{2} = v _{i}^{2} + 2gh\]

The velocity of the left side is the final velocity, the one on the right the initial velocity.

g = -32ft/s^2. h is the max ht if vsub f equals 0.

Answer 6

what is the small f and i

Answer 7

what do u mean by s^2

Answer 8

i still need help

Answer 9

small f = final...small i = initial

Answer 10

so how could i figure this question out

Answer 11

the question is in a attachment

Answer 12

Someone help me solve this equation

Answer 13

wootwoot

Answer 14

Hint:

g = 32
h = 0
v = 61

input and solve for t

Answer 15

do i use the current furmula they gave me in the math problem

Answer 16

Um, let me think for a minute......YES OF COURSE! :D

Answer 17

what do i put for t

Answer 18

You don't put anything for t. You solve for t.

Answer 19

or do i just leave it alone

Answer 20

Do you know what it means to solve for a variable?

Answer 21

yeah

Answer 22

H=-1/2 32t^2+61t+0

Answer 23

That's not solving it for t :/ You should end up with "t=...."

Answer 24

plz help shane what do i do

Answer 25

what class is this for?

Answer 26

what is the vertex for the porabola?

Answer 27

this is for algebra

Answer 28

I dont understand why people are saying solve for t, you need to find the vertex. do you know how to do this?

Answer 29

some show me how to solve this step by step

Answer 30

If it's for algebra I would just graph it and determine the max height of the graph using a trace.

Answer 31

but I'm generally lazy...

Answer 32

?

Answer 33

this is the question

Answer 34

vertex = -b/2a this will give you a value for t, then plug in that value

Answer 35

i dont know how to figure that out

Answer 36

vertex = -v/(2(-1/2*g) what does this equa?

Answer 37

so -v/-g = v/g = 61/32
now plug in (61/32) into your formula with the given value for v and g

Answer 38

im so counfused

Answer 39

You guys are overloading her with methods to solve the same problem/

Answer 40

From start to finish:
Solve for t as we were trying to get you to do. To do that all you need to do is solve this quadratic for t:
\[0=-16t^2+61t+0\]Use the quadratic formula or wolfram to see what t will be: http://www.wolframalpha.com/input/?i=Solve+for+t%3A+0%3D-16t^2%2B%2861%29t

Once you have t, half of t will be the point at which the height is at max. So plug (1/2)t back into the original equation and solve for H. That's it.

Answer 41

(-1/2)*g*t^2 + vt = (-1/2)*61*(61/32)^2 + 32(61/32) = 58.140

Answer 42

is that the answer

Answer 43

zzrock is that the answer

Answer 44

Yes...that's the answer. About 58ft...

Answer 45

is the nearest tenth 58.10

Answer 46

Yes

Answer 47

yay thank u guys sooooooooooooooooooo much

Answer 48

If you know how to use a graphing calculator you could have just graphed and had it tell you the max. 20 seconds of effort...again, I'm lazy :)

Answer 49

@zzr0ck3r: Finding the vertex of the parabola was definitely simpler but I guess we all just defaulted to finding t first and then solving for h.

Answer 50

yeah I hear ya. This is the way I learned to do it without calculus. Just because someimtes the way down and the way up are not the same distance, so the solve for t method would not work so great.