A particle of 5kg is released from 250m high and it enters 3.5m into the earth.find the net force(in N) applied to the particle by earth.The particle was primarily at rest.
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in the air : V^2=Vo^2+2gS since; the particle was primarily at rest therefore; Vo =0 therefore; V^2=0+2 9.8 250=4900 therefore; V=70 m/s . in the earth : since; Vo=70 m/s and V=0 a=-a(deceleration"as it is inside the earth") therefore; 0=4900-2 a 3.5 therefore; a=700 m/s^2. F=m(a+g)=5(700+9.8)=3549 N