A student uses a spring to launch a marble vertically in the air . the mass of the marble is 0.002kg and when the spring is stretched 0.05m it exerts a force of 20N. what is the maximum height the marble can reach ?
from the equestion: F=mh÷g 20N=0.002×0.05÷10 let make pair subject: hm=Fg 0.05m×0.002=20N×10 0.05m×0.002=200 0.05m=200-0.002=199.98 approximately 200. 0.05m÷0.05m=200÷0.05 m=4000 therefore; Maximum height is 4000m.
