Question

A slingshot obeying Hooke’s law is used to launch pebbles vertically into the air. You observe that if you pull a pebble back 11.2 cm against the elastic band, the pebble goes 5.7 m high.

A)Assuming that air drag is negligible, how high will the pebble go if you pull it back 22.4 cm instead?

B)How far must you pull it back so it will reach 11.4 m ?

C)If you pull a pebble that is twice as heavy back 11.2 cm , how high will it go?

Answer 1

this as problem based on conservation of energy

energy stored in sling= potential energy

1/2*K*X² = M*g*h

given that

you pull a pebble back 11.2 cm against the elastic band, the pebble goes 5.7 m high

1/2*K*(0.112)² = M*10*5.7

hence we get

K/M = 9088.01

a)when it pulls 22.4 thais twic of given

hence height will become 4 times

h =5.7*4 =22.8 meters

b) to reac 11.4 meter that is 2 times of given height it must pull the slin $\sqrt{2}times$

hence it should pull = 15.83 cm

c) if it's twice heavy it's height will become half

hence it will go only 5.7/2 = 2.85 meter