"If a ball of mass M is dropped from a height h onto a spring with spring constant k (whose equilibrium positions is at height 0), compresses the spring an additional distance (L/2), and then rebounds, what height will the ball reach? Express your answer symbolically."

So I was thinking something along the lines of h= M*k*(L/2) but I'm not sure :/ Any guidiance?

 h ’= k  (h +L/2)²/ (2 M g )

Explanation:

For this exercise, one of the best methods to solve it is with energy conservation.

Starting point. Lower, spring with maximum compression

              Em₀ = Ke = ½ k x²

Final point. Higher after bounce

              = U = M g h ’

             Em₀ = Em_{f}

             ½ k x² = M g h’

the compressed distance is

          x = h+ L / 2

where h is the distance that compresses the spring by the height where it comes from and L/2 the additional compression

        h ’= ½ k x² / M g

we calculate

       h ’= k  (h +L/2)²/ (2 M g )

       h' = (k/2Mg) h2 (1 + L/2h)2

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2