A rock is thrown upward with a velocity of 28 meters per second from the top of a 23 meter high cliff, and it misses the cliff on the way back down. When will the rock be 4 meters from the water, below?

Question

anonymous · Answer

Will some one please help me

whpalmer4 · Answer

Do you know the general formula$$h(t) = -\frac{1}{2}gt^2 + v_0t + h_0$$for the height of an object launched with initial velocity $v_0$ and initial height $h_0$?

whpalmer4 · Answer

Set that equal to the height of the rock when it is 4 meters from the water and solve for the positive value of $t$.

whpalmer4 · Answer

$$g=9.8\text{ m/s}^2$$

anonymous · Answer

i am sorry the numbers are 11 meters per second and 24 meter high cliff and when will the rock be 8 meters from the water

anonymous · Answer

i have no idea how to do this

whpalmer4 · Answer

Exactly the same way as if the numbers were 28 m/s, 23 meter high cliff, 4 meters from the water.

anonymous · Answer

ok i don't know how to set it up or solve it, i am really bad at algebra

whpalmer4 · Answer

Well, baby steps.  First, identify the variables that correspond to your numbers.

anonymous · Answer

ok

anonymous · Answer

h=8
g=9.8
v0=11m/s
h0=24

anonymous · Answer

right?

whpalmer4 · Answer

yes!

anonymous · Answer

ok then what?

whpalmer4 · Answer

so write out the formula

anonymous · Answer

well there are t's in the formula but i don't have number for those

whpalmer4 · Answer

Yes, we need to have a variable to solve for :-)

whpalmer4 · Answer

so just write $t$ in the appropriate spots for $t$

anonymous · Answer

8=1/2(9.8)t^2+11t+8

anonymous · Answer

right?

whpalmer4 · Answer

Nope.  Go back and look at my equation carefully.  There are at least two errors in your formula

anonymous · Answer

h=-1/2

anonymous · Answer

but im not sure where the other error is?

whpalmer4 · Answer

Also, you should put the 1/2 as (1/2) to make it crystal clear that it is $$-\frac{1}{2}gt^2$$and not$$-\frac{1}{2gt^2}$$

whpalmer4 · Answer

Okay, the missing negative sign was one error.  Can you please type the whole equation as you think it should be now?

anonymous · Answer

ok

anonymous · Answer

8=(-1/2)(9.8)t^2+11t+8

anonymous · Answer

right?

whpalmer4 · Answer

No.  I'm curious about the two 8's in the formula.  Wouldn't they just cancel each other out?

whpalmer4 · Answer

Can you tell me why 8 appears twice in the formula?

anonymous · Answer

well i was kind of confused with the h0 and just the h

whpalmer4 · Answer

okay.  let's look at my original formula:

$$h(t) = -\frac{1}{2}gt^2+v_0t+h_0$$$h(t)$ is the height at time $t$
$h_0$ is the initial height, as I mentioned when I wrote the original formula

Which, if either, of those should be 8 in your formula to find the time at which the rock is 8 meters above the surface of the water?

anonymous · Answer

i don't think it should be in the equation? right

whpalmer4 · Answer

What is the initial height of the rock?

anonymous · Answer

0?

whpalmer4 · Answer

Isn't it being thrown off a cliff?

anonymous · Answer

yah

anonymous · Answer

24meters high

whpalmer4 · Answer

Okay, so $$h_0 = 24$$What do we have for a formula now?

anonymous · Answer

8=(-1/2)(9.8)t^2+11t+24

anonymous · Answer

right?

anonymous · Answer

now what?

whpalmer4 · Answer

Yes.  Now use the quadratic formula to solve that for $t$.

If $$ax^2+bx+c=0,\,a\ne0$$$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

anonymous · Answer

−0.561224489796+1.125459783873i,−0.561224489796−1.125459783873i

anonymous · Answer

so how many secs is that?

whpalmer4 · Answer

no, you must have done the arithmetic incorrectly.  what did you use for a, b, c?

anonymous · Answer

i don't know my teacher gave me a website to type in a an equation and it tells you the answer for the quadric formula

anonymous · Answer

and it is usually right

whpalmer4 · Answer

what did you type in?

whpalmer4 · Answer

I'm betting it is always right if you type the formula correctly :-)

anonymous · Answer

8=(-1/2)(9.8)t^2+11t+24

anonymous · Answer

bc i have put stuff in there like that and it gives me the right answer

anonymous · Answer

what is it suppose to be

whpalmer4 · Answer

well, for starters, that isn't in the appropriate form, is it?

I said the quadratic formula applies if you have a function $$ax^2+bx+c=0$$
Is your equation in that form?

what is the url of this website?

whpalmer4 · Answer

I must say I'm not in favor of students not knowing how to solve quadratic equations with the formula without computer assistance!

anonymous · Answer

well how do i get in to that format?

whpalmer4 · Answer

Using the computer to do the drudge work once you've learned how to do it yourself is a completely different matter.

Do you have everything on one side, and 0 on the other?  If not, do some basic algebra until you do.

anonymous · Answer

0=(-1/2)(9.8)t^2+11t+24-8

whpalmer4 · Answer

yes.  Can you simplify that at all?

anonymous · Answer

0=(-4.9)t^2+11t+16

whpalmer4 · Answer

Good.  so what are the values of $a,b,c$ to go into the quadratic formula?

anonymous · Answer

a=-4.9t^2
b=11t
c=16

whpalmer4 · Answer

no, $a,b,c$ are just the coefficients, the variable doesn't come along for the right

anonymous · Answer

what????

whpalmer4 · Answer

if $a=-4.9, b=11, c=16$, then what is $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$just write it out first and show me, don't try to evaluate it

whpalmer4 · Answer

"come along for the ride" was what I meant to type