A rubber-band slingshot shoots a 25 g stone. When answering 2a below, remember that Ep = ½ kx2 and, because F = kx, Ep = ½ Fx, you will need to use the average force which is obtained by dividing the maximum force by two.

a. What is the initial speed of the stone if the rubber band is drawn back 0.15 m with a maximum force of 27 N?
b. How high will the stone rise if it is shot straight upward?

Question

A rubber-band slingshot shoots a 25 g stone.  When answering 2a below, remember that Ep = ½ kx2 and, because F = kx, Ep = ½ Fx, you will need to use the average force which is obtained by dividing the maximum force by two. 

a.	What is the initial speed of the stone if the rubber band is drawn back 0.15 m with a maximum force of 27 N?
b.	How high will the stone rise if it is shot straight upward?

anonymous · Answer

$$a.)Ep=1/2FX $$

no loss of energy so all the elastic energy will be converted into kinetic 
$$1/2FX =1/2MV ^{2}$$
solve for V
V=$$\sqrt{F*X/M}$$=12.728
B. all the kinetic energy gets converted into potential energy so
$$1/2MV ^{2}=MGH$$
Solve for H
$$H=V ^{2}/(2*G)$$=8.260 m

anonymous · Answer

or H=(1/(2g.m))*F*x