A ball is dropped from 27.0 ft above the ground. Using energy considerations only, what is the velocity of the ball just as it hits the ground? If the ball is thrown back up to a ledge 15 m above the ground, what would be the velocity just as it reaches the ledge?
Help with physics queston.
the potential energy at the start is mgh
That should equal the kinetic energy 1/2 mv^2 at the bottom.
On the last, assuming it was thrown up to reach 27ft, then
original energy=newpeat15 + KE at 15
mg*27=mg15+ 1/2 m v^2
To find the velocity of the ball just as it hits the ground, we can use the principle of conservation of energy. The formula for the potential energy is given by:
PE = mgh
where PE is the potential energy, m is the mass of the ball, g is the acceleration due to gravity, and h is the height.
When the ball is dropped from a height of 27.0 ft above the ground, its potential energy is converted entirely into kinetic energy just before it hits the ground. Therefore, we can equate the potential energy at the initial height to the kinetic energy just before it hits the ground:
PE = KE
mgh = (1/2)mv^2
Where v is the velocity of the ball just before it hits the ground. Mass cancels out, and we can solve for v by rearranging the equation:
v = sqrt(2gh)
Substituting the given values, we have:
v = sqrt(2 * 9.8 m/s^2 * 27.0 ft * 0.3048 m/ft)
Simplifying the calculations, we find the velocity just as the ball hits the ground.
Now, let's move on to the second part of the question.
To find the velocity just as the ball reaches the ledge, we can again use the principle of conservation of energy. The total mechanical energy (sum of potential and kinetic energy) of the ball is conserved throughout its motion.
Initially, the ball is thrown back up to a height of 15 m above the ground. At this height, the potential energy is given by:
PE = mgh
where PE is the potential energy, m is the mass of the ball, g is the acceleration due to gravity, and h is the height.
When the ball reaches the ledge, all of its potential energy is converted into kinetic energy. Therefore, we can equate the potential energy at the height of the ledge to the kinetic energy just before it reaches the ledge:
PE = KE
mgh = (1/2)mv^2
Again, mass cancels out, and we can solve for v by rearranging the equation:
v = sqrt(2gh)
Substituting the given values, we have:
v = sqrt(2 * 9.8 m/s^2 * 15 m)
Performing the calculation, we find the velocity just as the ball reaches the ledge.