Question

how do you solve this: a ball falls to the ground at a rate of sqrt(2h/g) if dropped from 10m how long does it take to come to rest.
g=9.8
ans:sqrt(20/g)(1+sqrt(p))/(1-sqrt(p))

Answer 1

Do you use the 5 Kinematic equations here?

Answer 2

not a physics problem. calc2.

Answer 3

@experimentX

Answer 4

oh. :(

Answer 5

ya, that's definitely my face!

Answer 6

Integrate the equation in h from ten to zero

Answer 7

where does the (1+sqrt(p))/(1-sqrt(p)) come in though?

Answer 8

perhaps sqrt(2hg)

Answer 9

the book says sqrt(2h/g)

Answer 10

ok ok ... what is p?

Answer 11

it doesn't say!!!

Answer 12

Here's the equation: \[\int\limits_{10}^{0} \sqrt{2h \div g}\]

Answer 13

the velocity is given as a function of distance. find the average velocity.

Answer 14

Let me put in g.\[\int\limits_{10}^{0}\sqrt{h \div 4.9}\]

Answer 15

That's simplifying the 2/9.8

Answer 16

divide the whole integration by 10. that will give you average velocity.
\[ \text{avg vel = }\frac{1}{10} \int_0^{10} \sqrt{\frac{h}{4.9}}dh\]

Answer 17

That doesn't really do anything.

Answer 18

i get that numb3r1, i need to find out what p is doing in this equation.

Answer 19

Let's walk through the integration instead.

Answer 20

you both do it your own ways.

Answer 21

looks like the answer is what numb3r1 got

Answer 22

i get how to integrate, this i already did that.

Answer 23

.451754sqrt(h)

Answer 24

\[\int\limits_{10}^{0} \sqrt{\frac{ h }{ 4.9 }}dh\] That should not be in the format sqrt (h). The integral of sqrt(h) dh is in sqrt (h^3)

Answer 25

and please excuse my pissed off behavior it's 314 in the morning.

Answer 26

No worries, it's past midnight here too.

Answer 27

OK, I'll just do the integral.

Answer 28

Ha. Sleep. I use to go on 4 hr sleep a day for a yr.

Answer 29

there shouldn't be 'p' ... perhaps you missed something in the question or typo on answer.

Answer 30

\[\int\limits_{10}^{0}\sqrt{\frac{ h }{4.9 }} dh={\frac{ 1 }{4.9 }} \int\limits_{10}^{0}\sqrt{h}\]

Answer 31

Sorry, that's wrong.

Answer 32

It should be root(4.9)

Answer 33

what the?? can you do that?

Answer 34

AH. better.

Answer 35

The root(4.9) is just a constant. Constants can be put in or taken out of integrals

Answer 36

sure, i know that much.

Answer 37

2x^(3/2)/3

Answer 38

h is x

Answer 39

\[\int\limits_{10}^{0}\sqrt{h}dh=\int\limits_{10}^{0}h^{1/2} dh\]

Answer 40

i've already gone ahead of you.

Answer 41

So yes, 2x^(3/2)/3 (then put back the 4.9)

Answer 42

Erm... root 4.9

Answer 43

Substitute in 10 for h, then subtract the same thing with 0.

Answer 44

root 1/4.9

Answer 45

Yeah, that.

Answer 46

sure, easy enough.

Answer 47

OK, what does that get you?

Answer 48

9.523

Answer 49

appx

Answer 50

OK, let's try not simplifying out g. In that case, you extract (sqrt 2/g) from the equation.

Answer 51

sure. that leaves sqrt2/g (21.082) appx

Answer 52

Can you check the answer to make sure you typed it in correctly?

Answer 53

and yes it's all correct.

Answer 54

Solve for p and see if it's anything useful?

Answer 55

huh???

Answer 56

This question... calculus problem or not, something seems off.

a rate of sqrt(2h/g)

hmm, rate implies a velocity in m/s... but that equation does not give you velocity, it gives you time for an object to fall. You can see here: http://www.gravitycalc.com/
\[\Large t = \sqrt { \frac{ 2h }{ g } }\]
You can confirm this by looking at the units of each (h is in metres, g is m/s^2):
\[\Large t = \sqrt { \frac{ m }{m s^{-2} } } = \sqrt{s^2} = seconds\]
Velocity IS given by \[\Large v = \sqrt { 2g h }\]
And this too "how long does it take to come to rest" ... I guess that's supposed to mean "how long does it take to hit the ground"? Or are we supposed to assume it bounces when it hits the ground, and find the time it takes to stop bouncing...?

Answer 57

it bounces.

Answer 58

trust me i've been doing this problem for about 2 weeks, and it still doesn't make any sense.

Answer 59

Well that seems like the fault of the question, it's a poorly worded/constructed problem... I have to guess p is a measure of the springyness of the ball.

Answer 60

As @agent0smith says the rate that the problem states is really the time to hit the ground. The problem is poorly constructed.

Answer 61

yes, i would have to agree with you there.

Answer 62

well, i paraphrased the question but there's no mention of p anywhere in the problem.

Answer 63

If those are the bounces, we have to find a way to sum up the time for each one.

http://www.sosmath.com/calculus/geoser/bounce/bounce.html