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A slingshot obeys Hooke's law and is used to launch pebbles vertically into the air. You observe that if you pull a pebble back 20 cm against the elastic
band, the pebble goes 6 m high. Assuming that air drag is negligible, if you pull a pebble that is twice as heavy back 20 cm, how high will it go?
4 m
5 m
3 m
6 m

Sagot :

ajeigbeibraheem

Answer:

24 m

Explanation:

By applying the law of conservation of energy.

K.E = P.E

[tex]\dfrac{1}{2}kx^2 = mgh[/tex]

By rearrangement;

[tex]\dfrac{k}{m} = \dfrac{2*g*h}{x^2}[/tex]

where;

k = spring constant

m = mass of pebble

h = height reached

x = stretched length

[tex]\dfrac{k}{m} = \dfrac{2*9.8*6}{(20 \times 10^{-2})^2}[/tex]

[tex]\dfrac{k}{m} =0.2940 \times 10^4[/tex]

Since the drag force is negligible;

[tex]\dfrac{1}{2}kx^2 = mgh[/tex]

[tex]h = \dfrac{k}{m}\dfrac{x^2}{g}[/tex]

[tex]h = \dfrac{1}{2}\times 0.2940 \times 10^4 \times \dfrac{(40\times 10^{-2})^2}{9.8}[/tex]

h = 24 m

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