"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

30 miles

step-by-step explanation:

we know that 60 seconds equals 1 minute and 60 minutes 1 hour

44 ft       60 sec     60 min     158400   ft

  *   *   =

1 sec       1 min       1 hour         hour

we know that there are 5280 ft per mile

158400   ft       1 mile         30 miles

  *   =

1   hour               5280 ft     1 hour

so in 1 hour   it can travel 30 miles

in if you ask questions, the points are taken from your total and given to the one who answered it..

so if i have 30 points and i ask a question for 5 points, then when someone asks a question i will lose 5 points and the answerer earns 5

happy to