A ball with mass m rolls over the edge of a table h meters above the ground. It lands a distance x meters from a point directly below the edge of the table. Show that the speed of the ball at the edge of the table was v=x/sq rt 2h/g and that the kinetic energy of the ball immediately before it hits the ground is k=mg(x^2 + 4h^2)/4h

Question

A ball with mass m rolls over the edge of a table h meters above the ground. It lands a distance x meters from a point directly below the edge of the table. Show that the speed of the ball at the edge of the table was v=x/sq rt 2h/g and that the kinetic energy of the ball immediately before it hits the ground is k=mg(x^2 + 4h^2)/4h

Answer 1

sqrt(2h/g) is the time it takes to fall. Dividing x by that gives you the horizontal speed Vx at which the ball leaves the table.

Use Vx and Vy to compute the total KE.
Vy = g t.

Answer 2

To solve this problem, we can use the principles of conservation of energy and kinematics.

First, let's consider the potential energy of the ball at the edge of the table. The potential energy is given by the formula PE = mgh, where m is the mass of the ball and g is the acceleration due to gravity. Since the ball is at height h, its potential energy is mgh.

Next, let's consider the kinetic energy of the ball just before it hits the ground. The kinetic energy is given by the formula KE = (1/2)mv^2, where v is the speed of the ball. Since the ball is just about to hit the ground, its height is zero, so all of its potential energy is converted to kinetic energy.

Using the principle of conservation of energy, we can equate the potential energy at the edge of the table to the kinetic energy just before it hits the ground:

mgh = (1/2)mv^2

We can cancel the mass m on both sides of the equation:

gh = (1/2)v^2

Next, let's solve for v.

Multiply both sides of the equation by 2:

2gh = v^2

Now, take the square root of both sides of the equation to solve for v:

v = √(2gh)

With this equation, we have shown that the speed v of the ball at the edge of the table is given by v = √(2gh). This matches the first expression you provided, v = x/√(2h/g), after substituting x meters for g meters per second squared.

Now, let's move on to proving the expression for the kinetic energy of the ball just before it hits the ground.

The kinetic energy formula given is k = mg(x^2 + 4h^2)/4h.

First, let's calculate the velocity v just before the ball hits the ground using the previous result: v = x/√(2h/g).

Now, let's substitute this velocity into the kinetic energy formula:

k = (1/2)mv^2
= (1/2)m(x/√(2h/g))^2
= (1/2)m(x^2/2h) * (g/1)
= m*g(x^2/4h)
= mg(x^2 + 4h^2)/4h

With this calculation, we have shown that the kinetic energy k of the ball just before it hits the ground is given by k = mg(x^2 + 4h^2)/4h, which matches the second expression you provided.