- #1

stephanna

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## Homework Statement

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.]

## 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

## 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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