A ball dropped from 20.0 ft above the ground....?

Question

a) At what height is half of its energy kinetic and half potential?

b) Using energy considerations only, what is the velocity of the ball just as it hits the ground?

Answer 1

a) You have dropped the ball from 20.fts

1ft = 0.3 meters

initial height = 6 meters

Initial energy = Potential energy = m*g*20

The energy at som height = m*g*h + 1/2*m*v^2

Now, using conservation of energy :

m*g*6 = m*g*h + 1/2*m*v^2

and if they should be half of its energy kinetic and halg potential

2m*g*h = m*g*6

h = 3 meters = 10ft

answer for a) = 10 fts = 3 meters

b) Using conservation of energy :

Initial energy = m*g*6

final energy = 1/2*m*v^2

1/2*m*v^2 = m*g*6

v = 10.8 m/s

Hope that helps

Answer 2

It is unfortunate that this problem is not in the metric system since the units are easier to handle. Nevertheless, the potential energy is PE = mgh where m is the mass of the object, g is the acceleration due to gravity (32 ft/s^2) and h is the height. mg is also equal to the weight of the object. Thus the PE in your problem is 20 x mg. The kinetic energy is 1/2 m v^2 where v is the velocity and is zero at the objects maximum height.

As the object falls, the PE decreases and the KE increases when the PE is one-half its initial value, the height must be 10 ft. The KE equals 10 mg. Energy is conserved during the fall.

When the object just hits the ground, the PE is totally converted to KE thus 20mg = 1/2 m v^2. The mass cancels out of this equality and we xolve for v. v = sqrt(40 g) or Sqrt(40 x 32) which you can evaulate using your calculator.

Hope this helps, Mike R

Answer 3

Dropping a ball of mass (m), from a height (h) of 20' above the ground.

Conservation of Energy:
Gravitational Potential Energy (GPE) = Kinetic Energy (KE)

GPH = mgh

The instant before it hits the ground, it has no GPE. All energy has been converted to kinetic energy, (KE).

KE = ½mv²

Part a)
At what height does mg(20-h) = ½mv²

The GPE is linear from zero to 20 feet. Since the energy is conserved, the sum of the GPE and KE must be constant. The only height where the two values could be equal is 10'

Part b)
GPE = KE
mg20 = ½mv² (cancel out the mass, m)
g20 = ½v²
v² = 40g
v = √40g; where g = 32 ft / sec²
v = 35.8 ft / sec

Answer 4

see you want the formula for kinetic power that = a million/2 mv sq (m= mass , v= velocity ,velocity) formula of power= w/t (artwork/time) potential power = mgh ( mass*gravitational stress*proper evaluate gravitational stress to be 10m/s) (s = distance) this all you want i will't provide the solutions bcz its a quite lengthy exaaaaaammmple :)

Answer 5

A) 10ft above the ground

B) about 7 feet per second(6.66666666666666 to be exact(it goes on and on))

Answer 6

mgh=.5mv^2

mas cancels leaving v and g. g = 9.81 m/s^2