A 70 kg person is at the top of a 10 m tall slide. Ignoring friction, what is his or her speed at the halfway point where he or she is 5 m from the bottom of the slide? What is his or her speed at the bottom of the slide?

Answer 1

V^2 = Vo^2 + 2g*d = 0 + 19.6*(10-5)=98
V = 9.90 m/s.

V^2 = 0 + 19.6*10 = 196
V = 14 m/s.