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31 October, 17:40

A 3.00 kg stone is dropped from a 39.2 m high building. when the stone has fallen 19.6 m, the magnitude of the impulse the earth has received from the gravitational force exerted by the stone is

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  1. 31 October, 20:43
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    The formulas are, Impulse = mv-mu ... (1) v^2 = u^2 + 2as ... (2) We know that, u=0 a=acceleration=gravity = 9.80665 m/s^2 = 9.81 m/s^2 s=19.6 sub (2) we get, v^2 = 0 + 2*9.81*19.6 v^2 = 2*9.81*19.6 v^2 = 384.552 v = 19.6099 v = 19.61 m/s Sub v=19.61 m/s in (1) we get, Impulse = mv - mu we know that u=0; v = 19.61 m/s; m = 3.00 kg Impulse = 3 (19.61) - 3 (0) Impulse = 58.83-0 Impulse = 58.83 Ns. Therefore the gravitational force exerted by the stone is 58.83 Ns.
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