Can anyone answer part 3 for me? I\'ve tried so many incorrect answers that I\'v

Question

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

Explanation / Answer

m g h1 = 1/2 k x1^2 conservation of energy m g h2 = 1/2 k x2^2 (I) h2 = (x2^2 / x1^2) h1 Using h1 = 3 for x1 = 2 then for x2 = 3 h2 = 9 * 3 / 4 = 6.75 m if you pull the slingshot 3 cm Now using 2m in equation (I) and dividing h2 = (x2^2 / x1^2) h1 / 2 where x1 = x2 = 3 and h1 = 6.75 h2 = 6.75 / 2 = 3.38 m

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