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A 0.11 kg ball is thrown straight up into the air with an initial velocity of 26 m/s
Find the momentum of the ball halfway to its maximum height on the way up.

V0 = 26 m/s
g=9.8 m/s^2

Max Height = V^2/2g
= 26^2 / 2*9.8
= 676 /19.6
= 34.489 meters

Halfway up is 1/2 this number or 17.2445 meters

Use the equation for height in terms of time
and solve for t

Height = Vt - 1/2 gt^2
17.2445 = 26t -1/2 (9.8)t^2
4.9t^2-26t+17.2445=0

Use the quadratic formula

( -b +- SQRT(b^2-4ac))/2a

to solve for roots

Roots are 0.777(ball on way up) and 4.52(ball on way down)

If you have an Ultracapacitor with the below specs…

Boostcap® Ultracapacitor
3000.0 Farad ± 20%; 2.7V
How many would you need to store 1KWH of energy? ( 1,000 watts for an hour)

The energy in Joules stored in a capacitor can be calculated with the following formula

E=(1/2) C V^2

for C=3000F V=2.7 Volts

E=10935 Joules

1Kwh = 3,600,000 Joules

Therefore you would need

3600000/10935 = 329.2 or rounding up

330 capacitors to supply 1Kwh of energy

Modern ultracapacitors are now approaching the energy density of the best batteries and have the advantage of very rapid charge and discharge currents

The device shown at the following link is very similar to the calculated result shown above
Capacitor Bank

A total of 10^4 kg of water per second flows over a waterfall 25m high!
If half of the power this flow could be converted into electricity, how many 100W light bulbs could be supplied?

This is a classic gravitational potential energy problem where we are asked for electrical power.

The formula for the potential energy of the water is
P=mgh

where
m = 1×10^4 Kg
g=9.8m/s^2
h=25m

the power is simply the amount of energy available in one second or:

P= 2.45 x 10^6 J/s

With 50% efficiency

Power = 1.225 x 10^6 Watts
divide by 100 watts/bulb
12,250 light bulbs

A 500g rubber ball is dropped from a height of 10m and undergoes a perfectly elastic collision with the earth.

(a) What is the Earth’s velocity after the collision, assuming the Earth is at rest just before the collision.
(b) How many years would it take the Earth to move 1.0mm at this speed?

Well, the second assumption in statement (a) is incorrect, but let’s do the math anyway:

(a)First calculate time for ball to drop 10m
s=1/2 g t^2

s=10m
g=9.8 m/s^2 (the accelaration due to gravity)

solve for t

t= sqrt(2s/g) = sqrt (2.04) or about 1.43 seconds