"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 guidance?

Step-by-step explanation:

Given

Mass of ball is M  is dropped from a height h

spring constant is k

Here we can use conservation of energy

Mass has a Potential energy of E=mgh

and when it falls on the spring then it compresses it a length of 0.5 L

Now potential energy is converted into Elastic potential energy and then spring unleashes this energy to the mass and again provide kinetic energy to mass to move upward .

In this way energy is conserved. So, mass must move to an initial  height h

considering

∈3,600

step-by-step explanation:

false

step-by-step explanation:

the reason i say this is because for a certain length of sides you can pick any two and the length is wither equal or greater than the third side and in this case it's not.

3 + 5 = 8

8 is smaller than 9 which is the third side.

hope this : )