Answer:
The distance from the edge of the table at which the ball will hit the ground is approximately 0.49 meters
Explanation:
The mass of the ball = 0.84 kg
The spring constant k = 142 N/m
The compression of the spring = 0.06
Therefore, by conservation of energy, we have;
The potential energy of the spring = The kinetic energy given to the ball
The potential energy of the spring = 1/2 × k × x² = 1/2 × 142 × 0.06² = 0.2556
The potential energy of the spring = 0.2556 J
The kinetic energy given to the ball = 1/2 × m × v²
Where;
v = The final velocity of the ball
Substituting gives;
1/2 × 0.84 × v² = 0.2556
∴ v = √(0.2556/(1/2 × 0.84)) ≈ 0.78
The final velocity of the ball = v ≈ 0.78 m/s
The height of the ball above the ground, h = 2.0 m
Therefore, we have;
The time, t, it takes the ball to touch the ground from the 2.0 m height, h, is given as follows;
h = 1/2·g·t²
t = √(2·h/g)
Where;
g = The acceleration due to gravity = 9.81 m/s²
∴ By substitution, we have;
t = √(2 × 2/9.81) ≈ 0.63855
t ≈ 0.63855 s
The distance from the edge of the table at which the ball will hit the ground = The horizontal distance covered during before the ball hits the ground
The horizontal distance covered during before the ball hits the ground = The (horizontal) velocity of the ball × The time it takes the ball to touch the ground
∴ The horizontal distance covered during before the ball hits the ground = 0.78 × 0.62855 ≈ 0.490269 ≈ 0.49
The distance from the edge of the table at which the ball will hit the ground = The horizontal distance covered during before the ball hits the ground ≈ 0.49 m
