Asked by Jessica
A 3-kg stone is dropped from a height of 100 m. Find its kinetic and potential energies when it is 50 m from the ground.
Answers
Answered by
Henry
PE = mgh = 3 * 9.8 * 100 = 2940 J.
Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.6*50 = 980,
Vf = 31.3 m/s.
KE = 0.5m*V2 = 1.5*(31.3)^2 = 1470 J.
PE = 2940 - 1470 = 1470 Joules.
Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.6*50 = 980,
Vf = 31.3 m/s.
KE = 0.5m*V2 = 1.5*(31.3)^2 = 1470 J.
PE = 2940 - 1470 = 1470 Joules.
Answered by
Jessica
Where did you get the 19.6 from?
Also what formula are you using? The Vf^2 = Vo^2 + 2g*d....i've never seen that before
Also what formula are you using? The Vf^2 = Vo^2 + 2g*d....i've never seen that before
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