# Other equations

• November 13th 2009, 07:33 AM
Paul46
Other equations
Hi,

I'm from over the PHF site, i was wondering if you masters of maths could help me or just say no that's it thats the equation to use?

Obviously by now i realise the asker should post their attempt, so i will, but what i'm asking is for an alternative solution.

Example:

A stone is dropped from a cliff in the last 2 secs it travels a decimal fraction of 0.5665 of the total height, find the total time?

This is my attempt of which i know will be correct, sorry but i haven't mastered laTEX as of yet!

2/(1-sq-rt(1-0.5665)) = 5.855 secs

I know how to click the thanks button too!(Rofl)
• November 13th 2009, 07:37 AM
stapel

The basic equation for this sort of exercise is:

$s(t)\, =\, -4.9t^2\, +\, v_0 t\, +\, h_0$

...where s is the height (in meters) at time t seconds, v_0 is the initial velocity, and h_0 is the initial height. You are given s(T) and s(T+2) = 0.

I'm not sure what is meant by "a decimal fraction of 0.5665 of the total time"...? Is this the exact text of the exercise?

Note: Since the object is "dropped", then v_0 = 0, so the equation reduces to:

$s(t)\, =\, -4.9t^2\, +\, h_0$
• November 13th 2009, 07:42 AM
Paul46
• November 13th 2009, 07:46 AM
stapel

I've given you that s(T + 2) = 0, so s(T) - s(T + 2) = s(T), which is also the height covered in the last two seconds. How far have you gotten in using this fact?

Please be complete. Thank you! (Wink)
• November 13th 2009, 07:56 AM
Paul46
The source of my equation was from a poster in PHF, i tried it out to find it correct, my reasoning is that i know it's right but need to know if it can be simplified. I've gotten far enough by the fact that it is correct.

Sorry, am i making myself look a fool here?

You have given me an alternative, thank you! this is what i was searching for!

What do you wish me to be complete about? have i not given enough info?
• November 13th 2009, 08:18 AM
HallsofIvy
As Stapel said, the "basic" equation here is $s(t)\, =\, -4.9t^2\, +\, v_0 t\, +\, h_0$, the height, in meters, after t seconds, of an object falling from height $h_0$ meters, with initial upward speed $v_0$ meters per second, and acceleration (due to gravity) of -9.8 meter per second per second.

You problem, if I read this correctly, is to find the time a rock drops from some initial height, if the last two seconds it goes 0.5665 if the total height.

Okay, since it "dropped" it was not thown upward or downward. Its initial speed is 0. Writing the total height as "h", in meters, its height after t seconds is $h(t)= -4.9t^2+ h$. Let T be the total time of fall- the number you are looking for. Then $h(T)= -4.9T^2+ h= 0$, since it is now at the bottom. Also, if it has to cover the last 0.5665 of the total height in the last two seconds, it must have gone (1- 0.5665)h by T-2: $h(T-2)= -4.9(T-2)^2+ h= (1-0.5665)h$ so that, subtracting h from both sides, $-4.9(T-2)^2= (1- 0.5665)h- h= -0.5665h$.

The equation $-4.9T^2+ h= 0$ is the same as $h= -4.9T^2$ and the equation $-4.9(T-2)^2= -0.5665h$ is the same as $h= (4.9/0.5665)(T-2)^2$. Since those are both equal to h, you can set them equal to each other and solve for T. Once you have T, use either one to find h.
• November 13th 2009, 08:34 AM
Paul46
Hallsofivy,

Well first of all thank you for answering me without taking a bad attitude to my question, if i was a BRILLIANT mathematician i wouldn't be asking! I once again thank you and for not trying to make me look an idiot. Your answers are both obviously good, but you explained yourself better of which makes you a better person.