Answer:
[tex]\dfrac{H}{2}[/tex]
Explanation:
For the first time, the ball is shot with velocity [tex]v_1[/tex]. This velocity comes from the spring which is given by
[tex]v = \omega\sqrt{A^2-x^2}[/tex]
[tex]\omega[/tex] 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,
[tex]v = \omega A[/tex]
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
[tex]v_f^2 = v_i^2 + 2as[/tex]
[tex]v_f[/tex] is the final velocity,
[tex]v_i[/tex] is the initial velocity,
a is the acceleration
s is the displacement.
Substituting with our parameters,
[tex]0^2 = (\omega A)^2 - 2gH[/tex]
[tex]H = \dfrac{(\omega A)^2}{2g}[/tex]
For the second shot of the ball, the spring is half compressed. Hence x is at half amplitude. Thus,
[tex]v = \omega\sqrt{A^2 - (A/2)^2} = \omega\dfrac{A}{\sqrt{2}}[/tex]
Using this value of v to calculate the new maximum height, [tex]H_2[/tex],
[tex]H_2 = \dfrac{(\omega\frac{A}{\sqrt{2}})^2}{2g}[/tex]
[tex]H_2 = \dfrac{(\omega A)^2}{2\times2g} = \dfrac{H}{2}[/tex]
