How long does a ball with a mass m take to hit the ground?


by stephanna
Tags: kenetic energy, potential energy, time, velocity
stephanna
stephanna is offline
#1
Dec2-09, 12:05 PM
P: 4
1. The problem statement, all variables and given/known data

At t = 0 a rubber ball of mass m is dropped from a height of 10 m.
(i) When does it reach the ground? How fast is it moving when it hits the ground?
Each time it bounces, the ball looses 20% of its energy.
(ii) Calculate the height it reaches on the 1st bounce, and the time at which it hits the
ground for the second time and for the third time.
(iii) Find a formula for the time of the nth bounce. When does the ball come to rest?
[Hint: Look carefully at the times between bounces. You may use any formulae you know
from basic mechanics. The acceleration due to gravity is g = 9:81 m s^-2.]

2. Relevant equations
K.E. =1/2 * MV^2 Where M=mass V= Verlocity G=Gravity H=HEIGHT T=Time
P.E. =MGH
SO ....GH=1/2 * V^2 ....M is cancelled from each side
V=H/T and V=(2*G*H)^-1/2 so both equations equal each other hence
T=H/(2*G*H)^-1/2
H=(1/2 * V^2)/G

3. The attempt at a solution
PART i)

GH=1/2 * V^2... T=H/(2*G*H)^-1/2

10/(2*10*9.81)^-1/2= 0.713921561 SECONDS

V=H/T 10/0.713921561 ms^-1 = 14.00714105

ii)
K.E. =1/2 * MV^2
1/2 * m *14.007^2 = 98.098MJ M is still in the equation as the mass is unknown

20% of the energy is now taken off

98.098MJ - ((98.098MJ/100)*20)= 78.4784MJ

V^2=(2*K.E.)/M
(78.4784MJ* 2)/M=156.9568

H=(1/2 * V^2)/G
H=(1/2*156.9568)/9.81= 8m Height at first bounce

Here is where i had the trouble
Im tring to find the time of the 2nd bounce but it works out smaller that the 1st bounce
20% of the energy is now taken off
78.4784MJ-((78.4784MJ/100)*20)= 62.7827MJ

V^2=(2*K.E.)/M
(62.7827MJ*2)/M= 125.56544

H=(1/2 * V^2)/G
H=(1/2*125.56544)/9.81= 6.3998m

T=H/(2*G*H)^-1/2 OR T= H/V 0.571125223
T=6.3998m/(2*9.81*6.3998)^-1/2 =0.571125223 seconds this value is smaller than the first value
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stephanna
stephanna is offline
#2
Dec2-09, 12:18 PM
P: 4
I have genuanly attempted this and have been working on it over and over for a number of hours with our any progression, I would apprechate any advice
MaxL
MaxL is offline
#3
Dec2-09, 12:23 PM
P: 60
The time for a bounce should decrease with each bounce.

Try it yourself! Find a ball and bounce it. (Ball bearings on hard metal work especially well: Tink...Tink..Tink tink tink tnktnktnktkkkkkkkk...)

stephanna
stephanna is offline
#4
Dec2-09, 01:04 PM
P: 4

How long does a ball with a mass m take to hit the ground?


[QUOTE=MaxL;2470621]The time for a bounce should decrease with each bounce.
QUOTE]

I know but due to my calculations it doesnt n I dont know where I have gone wrong
MaxL
MaxL is offline
#5
Dec2-09, 01:36 PM
P: 60
Oh! Okay, so right off the bat I see one big mistake, in your 'relevant equations' section.

V=h/t

well, that's only true if the velocity is constant. But the ball is accelerating! Also, it follows that T=H/(2*G*H)^-1/2 is not, in general, true.

You got the right first velocity, but it was a total coincidence. You also got the first time wrong.

ALSO, V=Sqrt[2*g*H] gives you the velocity right before the ball hits the ground, not at any other time in the fall. You probably already knew that, but I thought I'd check to make sure.

Hang in there! This is a really cool problem, and it's worth figuring out.
MaxL
MaxL is offline
#6
Dec2-09, 10:15 PM
P: 60
It seems like you need a bit more help. I'll explain the approach that I used.

First, let's figure out t1, the time at which the ball hits the ground for the very first time. This requires the equation relating distance to time in a constant acceleration situation, that is, d=1/2*a*t^2. Here, d is Ho, the height of the building. a is acceleration due to gravity, g. So put it together, and get t1.

Now on to higher t's. t2=t1+(the time it takes for the first bounce arc). I used a trick to get this. I realized that, if the ball loses 20% of its KE, then the next bounce wil only go .8 the height of the first bounce, so H1=.8Ho. Then I used the same process I used to get t1, except I also multiplied by 2 (because the ball has to go up to the top of the arc, then come back down).

And then on to t3! t3=t2+(time it takes for second bounce arc). Now the maximum height of the bounce is reduced by another 20%. Can you see where this is going?

Now for the velocities- you want the v right before the velocity of the ball hits the ground? Well, at that point, it will have zero potential energy (using the ground as baseline, of course), so all of the potential energy it had at H-whatever will be KE.

I hope that helps!
Cryxic
Cryxic is offline
#7
Dec3-09, 09:43 AM
P: 53
Wait this is weird.....is this a double post or something? I already answered this question at this other thread created by the same user: http://www.physicsforums.com/showthread.php?t=359872
MaxL
MaxL is offline
#8
Dec3-09, 09:58 AM
P: 60
Looks like a double post to me. Moderator?


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