How fast must a ball be rolled along a 100 cm high table so that when it rolls off the edge it will strike the floor at a distance of 100 cm from the point directly below the table edge? Please show work.
To find the speed at which the ball must be rolled along the table, we can use the principle of conservation of energy.
First, let's consider the energy of the ball at two points: when it is on the table and when it is in the air just before hitting the ground.
1. On the table: The ball initially possesses only gravitational potential energy (GPE). The formula for GPE is given by mass (m) times the acceleration due to gravity (g) times the height (h). In this case, the height is 100 cm, which is equal to 1 meter. The GPE while on the table can be expressed as GPE_onTable = m * g * h.
2. In the air: After the ball rolls off the edge of the table, it loses all its potential energy and starts gaining kinetic energy (KE). The formula for KE is given by half of the mass times the square of the velocity (v). Therefore, KE_inAir = 0.5 * m * v^2.
According to the conservation of energy principle, the potential energy on the table is equal to the kinetic energy in the air (assuming no energy is lost due to friction or air resistance). Therefore, we can set up the equation:
GPE_onTable = KE_inAir
m * g * h = 0.5 * m * v^2
The mass (m) cancels out on both sides of the equation, so we can rearrange it to solve for the velocity (v):
v^2 = 2 * g * h
Now, substitute the known values: the acceleration due to gravity (g) is approximately 9.8 m/s^2, and the height (h) is 1 meter. Plugging in these values, we get:
v^2 = 2 * 9.8 * 1
v^2 = 19.6
Finally, take the square root of both sides to solve for the velocity (v):
v = √19.6
Approximately, v ≈ 4.43 m/s
Therefore, the ball must be rolled along the table at a speed of approximately 4.43 m/s.