posted by . on .
Can anyone answer part 3 for me? I've tried so many incorrect answers that I've already given up: 1.00, 1.50, 2.00, 3.00, 9.40
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 2.00 cm against the elastic band, the pebble goes 3.00 m high.
Part A (already answered correctly): Assuming that air drag is negligible, how high will the pebble go if you pull it back 5.00 cm instead? Answer: h = 18.8 m
Part B (already answered correctly): How far must you pull it back so it will reach 13.0 m? Answer: x = 4.16 cm
Part C: If you pull a pebble that is twice as heavy back 3.00 cm, how high will it go? Answer: h = ___ cm
given 3m high, at 2cm pull.
OK, you are pulling it 50 percent more, and doubling mass.
doubling mass will halve the height (PE=mgh)
The PE in the elastic goes by 1/2 k x^2, so if you increase the x to 1.50, that squared is 2.25x, so
doubling mass, pulling to 3cm, should make height 3m*1/2*2.25