A ball is rolled off a table that is 0.481 m above the floor. The ball is rolling with a velocity if 1.8 m/s as it goes off the edge of the table. At the exact instant the first ball rolls off the table a second ball is dropped from the same height.

A.) How long does it take each ball to reach the ground?
B.) What is the velocity of each ball the instant before it hits the ground?

Looking for the answer and the work that provided the correct answer. Thank you so much!!

A. h = Vo*t + 0.5g*t^2 = 0.481 m.

0 + 4.9t^2 = 0.481
t^2 = 0.09816
t = 0.313 s. Each ball.

B. V = Vo + g*t
V = 0 + 9.8*0.313 = 3.07 m/s. Each ball.

To find the answers to these questions, we can use the equations of motion for uniform acceleration.

A) How long does it take each ball to reach the ground?

Let's consider the first ball that is rolling off the table:

The vertical distance traveled by the ball is the height of the table, which is 0.481 m.

Using the equation:
Distance = Initial Velocity * Time + (1/2) * Acceleration * Time^2

Here, the initial velocity is 1.8 m/s, and the acceleration is due to gravity, which is approximately 9.8 m/s^2 (assuming negligible air resistance).

Plugging in these values, we get:
0.481 = 1.8 * t + (1/2) * 9.8 * t^2

Rearranging the equation, we get a quadratic equation:
4.9t^2 + 1.8t - 0.481 = 0

Solving this quadratic equation using the quadratic formula or factoring, we find two possible solutions for time (t):
t ≈ -0.287 s (this is not a valid solution)
t ≈ 0.097 s

The time it takes for the first ball to reach the ground is approximately 0.097 s.

Now let's consider the second ball that is dropped from the same height:

For this ball, it is dropped vertically, so the equation becomes:
Distance = (1/2) * Acceleration * Time^2

Plugging in the values:
0.481 = (1/2) * 9.8 * t^2

Simplifying the equation, we get:
4.9t^2 = 0.481

Solving for time (t), we find:
t ≈ 0.303 s

The time it takes for the second ball to reach the ground is approximately 0.303 s.

B) What is the velocity of each ball the instant before it hits the ground?

To find the velocity of each ball just before it hits the ground, we can use the equation:

Final Velocity = Initial Velocity + Acceleration * Time

For both balls, the acceleration is due to gravity, which is approximately 9.8 m/s^2.

For the first ball:
Initial Velocity = 1.8 m/s
Time = 0.097 s

Plugging in these values, we get:
Final Velocity = 1.8 + (9.8 * 0.097)

Simplifying the equation, we find:
Final Velocity ≈ 2.67 m/s

The velocity of the first ball just before it hits the ground is approximately 2.67 m/s.

For the second ball:
Initial Velocity = 0 m/s (as it is dropped)
Time = 0.303 s

Plugging in these values, we get:
Final Velocity = 0 + (9.8 * 0.303)

Simplifying the equation, we find:
Final Velocity ≈ 2.98 m/s

The velocity of the second ball just before it hits the ground is approximately 2.98 m/s.

Therefore, the answers to the given questions are:
A) It takes approximately 0.097 seconds for the first ball to reach the ground and approximately 0.303 seconds for the second ball to reach the ground.
B) The velocity of the first ball just before it hits the ground is approximately 2.67 m/s, and the velocity of the second ball just before it hits the ground is approximately 2.98 m/s.