A 21 g ball is shot from a spring gun whose spring has a force constant of 579 N/m. The spring can be compressed 1 cm. How high will the ball go if the gun is aimed vertically? The acceleration of gravity is 9.81 m/s 2 . Assume that at the end of the barrel, the spring’s displacement and the ball’s height are each zero

The answer  to the question is

The ball will go 0.14 meters high if the gun is aimed vertically

Explanation:

The energy in the spring → Energy, E =

Where E = energy in the spring

k = Spring constant

x = Spring compression or stretch

Therefore E =

The spring energy is transferred  to the ball as kinetic energy based on the first law of thermodynamics which states that energy is neither created nor destroyed

Kinetic energy = KE =

From which v = = = 1.66 m/s

from v² =u² - 2·a·S

Where v = final velocity = 0 m/s

u = initial velocity = 1.66 m/s

a = g = Acceleration due to gravity

S = height

Therefore 0 = 1.66² - 2×9.81×S

or S = 1.66² ÷ (2×9.81) = 0.14 m

answer; equivalent to a one(1) horsepower motor;