A3.00 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

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.

This is the answer: body 1 + engine 2