A 7 kilogram bowling ball is dropped from a fourth story window. If each story is 3 meters high, use energy conservation (not the equations of motion) to find the ball's velocity when it strikes the ground. Ignore air resistance.
1/2 m v^2= mgh
v= sqrt (2g*(3+.5))
I am assuming the window on the 4th floor is halfway up.
To find the ball's velocity when it strikes the ground using energy conservation, we need to consider the changes in potential and kinetic energy.
The potential energy of an object at a certain height is given by the equation PE = mgh, where m is the mass (in kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (in meters).
In this case, the ball is dropped from the fourth story of a building, which is equivalent to a height of 4 stories x 3 meters/story = 12 meters. So, the initial potential energy of the ball is PE_initial = mgh = 7 kg x 9.8 m/s^2 x 12 m = 823.2 Joules.
As the ball falls towards the ground, its potential energy decreases while its kinetic energy increases, according to the law of conservation of energy. The kinetic energy of an object is given by the equation KE = 0.5mv^2, where m is the mass (in kg) and v is the velocity (in m/s).
At the moment the ball strikes the ground, all of its potential energy is converted into kinetic energy. So, setting the initial potential energy equal to the final kinetic energy, we have:
PE_initial = KE_final
mgh = 0.5mv^2
We can cancel out the mass (m) and solve for v:
gh = 0.5v^2
2gh = v^2
v = sqrt(2gh)
Plugging in the values, we get:
v = sqrt(2 x 9.8 m/s^2 x 12 m)
v ≈ 14.85 m/s
Therefore, the ball's velocity when it strikes the ground is approximately 14.85 m/s.