A spring-loaded toy gun is used to shoot a ball straight up inthe air. The ball reaches a maximum height H, measuredfrom the equilibrium position of the spring. The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the ball go this time?

H' = H/4

Explanation:

By applying the law of conservation of energy to this problem, we know that:

Elastic Potential Energy Stored by Spring = Gravitational Potential Energy of ball

(1/2)kx² = mgH

H = (1/2)kx²/mg   equation (1)

where,

H = Height reached by the ball

x = compression of spring

k = stiffness of spring

m = mass of ball

g = acceleration due to gravity

Now, if we make the compression to half of its value:

x' = x/2

then:

H' = (1/2)k(x/2)²/mg

H' = (1/4)(1/2)kx²/mg

using equation (1), we get:

H' = H/4

answer: good job it's all correct !

explanation: