Kenetric Energys, verolcity and mass of a falling object?

[stephanna]

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. Homework 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 canceled 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

 

[Cryxic]

For Part i, use the equation y = (1/2)gt^2, where y is the height and t is the time. Solve for t. You should get 1.43 seconds. To find out how fast it's moving, use the equation v = sqrt(2gh), where v is the speed and h is the height. You should get 14 meters per second.

For part ii, if it loses 20% of its energy on the first bounce, that means it still has 80% of it left. From the gravitational potential energy equation, U = mgh, you know that the new potential energy will be 0.8*U, and from there you can figure out the height. It should be 8 meters after the first bounce. Knowing this technique and the given hint, you can figure out the rest.

 

[kerimek]

iii)
T=H/(2*G*H)^-1/2
T=H/(2*9.81*H)^-1/2
T=((2*H)^1/2)/(2*9.81)^1/2
T=((2*H)^1/2)/(4.417))
T=H^1/2/4.417

Now we know that the height decreases by 20% each time
So the formula for the height is H=(0.8)^n * 10 where n is the number of bounces

So the formula for the time of the nth bounce is:
T=((2*(0.8)^n * 10)^1/2)/4.417
T=((1.6*(0.8)^n)^1/2)/4.417
T=(1.26*(0.8)^n)^1/2

The ball will come to rest when its height is 0, so:
(0.8)^n * 10 = 0
(0.8)^n = 0
n=0

Therefore, the ball will come to rest after the 0th bounce, which is when it is dropped from the initial height of 10m.