How fast must the ball rolled along the surface of a 70cm high table so that when it roll off the edge it will strike the floor at the same distance (70cm) from the point directly below the edge of the table.

h= g•t^2/2 => t =sqrt(2•h/g)= =sqrt(1.4/9,8)=0.38 s.

L =v(x) •t => v(x) = L/t =
=0.7/0.38 =1.85 m/s.

h= g•t^2/2 => t =sqrt(2•h/g)= =sqrt(1.4/9,8)=0.38 s.

L =v(x) •t => v(x) = L/t =
=0.7/0.38 =1.85 m/s.

To determine the required speed of the ball, we can use the principle of conservation of energy. The potential energy the ball possesses at the edge of the table will be converted into kinetic energy as it falls and rolls off the table.

Step 1: Calculate the potential energy at the edge of the table.
Potential energy (PE) is given by the formula:
PE = m * g * h
where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height of the table (70 cm = 0.7 m).

Step 2: Calculate the kinetic energy at the point of impact with the floor.
Kinetic energy (KE) is given by the formula:
KE = 0.5 * m * v²
where m is the mass of the ball (assuming it remains constant), and v is the speed of the ball.

Step 3: Equate the potential energy to the kinetic energy.
PE = KE
m * g * h = 0.5 * m * v²
Canceling out the mass (m) from both sides:
g * h = 0.5 * v²

Step 4: Solve for v.
v² = 2 * g * h
Taking the square root of both sides:
v = √(2 * g * h)

Step 5: Substitute the given values and calculate the speed.
v = √(2 * 9.8 m/s² * 0.7 m)
v ≈ √(13.72)
v ≈ 3.71 m/s

Therefore, the ball must be rolled along the surface of the 70 cm high table at a speed of approximately 3.71 m/s to strike the floor at the same distance (70 cm) from the point directly below the edge of the table.

To find the required speed at which the ball must be rolled along the surface of the table, we can use the principle of conservation of mechanical energy.

Here's how we can approach this problem:

1. Identify the known quantities:
- Height of the table, h = 70 cm
- Vertical distance traveled by the ball after rolling off, H = 70 cm

2. Determine the potential energy of the ball at the edge of the table:
- The potential energy at the edge of the table is given by PE = m * g * h, where m is the mass of the ball and g is the acceleration due to gravity (9.8 m/s^2).
- Since mass is not given, we can assume the mass cancels out in this particular problem, so we only need to consider the height.

3. Determine the kinetic energy of the ball just before it reaches the floor:
- The kinetic energy just before the ball reaches the floor is given by KE = (1/2) * m * v^2, where v is the velocity of the ball.
- At this point, the ball is at a height H from the floor, which is equal to the original height of the table.

4. Apply the conservation of mechanical energy:
- According to the conservation of mechanical energy, the total energy remains constant.
- This means that the potential energy at the edge of the table is equal to the kinetic energy just before reaching the floor, i.e., PE = KE.

5. Set up and solve the equation:
- Equating the potential energy and kinetic energy, we get:
m * g * h = (1/2) * m * v^2
- Canceling out the mass, the equation simplifies to:
g * h = (1/2) * v^2
- Rearranging the equation to solve for v, we get:
v^2 = 2 * g * h
v = sqrt(2 * g * h)

6. Substitute the values:
- Plugging in the known values, we find:
v = sqrt(2 * 9.8 m/s^2 * 0.7 m)

7. Calculate the result:
- Evaluating the expression, we get:
v ≈ 5.977 m/s

Therefore, the ball must be rolled along the surface of the table at a speed of approximately 5.977 m/s in order to strike the floor at the same distance from the point directly below the edge of the table.