---Spring Question--- A 0.84 kg ball is held at rest against a spring (k = 142 N/m) that has been compressed 0.06 m. The ball and spring are on a table and the ball is 2.0 m above the ground. If friction between the ball and the table is negligible, then how far from the edge of the table (in meters) will the ball hit the ground?

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

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s

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