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 samegun, but this time the spring is compressed only half as far beforefiring. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which thespring is compressed is negligible compared to H.

Explanation:

For the first time, the ball is shot with velocity . This velocity comes from the spring which is given by

is the angular velocity of the ball,

A is the amplitude or maximum displacement of the spring from the equilibrium

x is the displacement at any time measured from the equilibrium point

At the equilibrium point, x = 0. Hence,

This is the maximum velocity.

Consider the upward motion of the ball with this initial velocity. Because the ball is going upward, the acceleration of gravity is negative (this is a personal choice for convenience). At maximum height, final velocity is zero. We use the equation of motion

is the final velocity,

is the initial velocity,

a is the acceleration

s is the displacement.

Substituting with our parameters,

For the second shot of the ball, the spring is half compressed. Hence x is at half amplitude. Thus,

Using this value of v to calculate the new maximum height, ,

dang that balls went far