How high could a .5kg basketball be launched by a spring, if the spring is compressed by 0.2m and has a constant of 200N/m?
How fast would the basketball be going just as it left the spring?
The potential energy of the spring would be
PE=1/2 200 ( .2)^2
this could be translated to a height of..
1/2 200 *.02^2= .5 * 9.8*h solve for h.