A 58.5 kg child slides down a water slide with a velocity of 1.2 m/sec at the top. At the bottom of the slide, she is moving horizontally, y=1.5 meters above the water. She splashes into the water d=2 meters to the left of the bottom of the slide.
1) Assuming potential energy to be zero at the water level, what is the mechanical energy of the child at the top of the slide?
b) Find h
KE = .5*m*v^2
Wg = mgh
PE = Wc
Yf = Yo + Vo*t + .5*g*t^2
ΔKE = ΔPE
Vf = Vo + a*t
The Attempt at a Solution
So, this is what I tried, and it turned out to not be correct. Maybe I made a mistake somewhere and if so would someone be kind enough to point it out/ help me figure out the correct way to solve this. Thank you.
So I started with knowing there was some initial KE:
KEt = .5 * 58.5 * (1.2)^2 = 42.12 J
I knew also that at the bottom there is:
KEb = .5 * 58.5 * v^2 = 29.25 * v^2
I knew from the problem that at the bottom of the slide, the girl is not moving in the Y direction, thus I can assume a standard 2-D kinematic type of problem to solve for V
1.5 = .5 * 9.81 * t^2
t^2 = 3 / 9.81
t = .55 sec
V = 2m / .55sec = 3.64 m/s
I then knew the KE at the bottom when the girl hit the water to be
KE = 29.25(3.64)^2 = 386.78 J
From Ei = Ef I said that the MEo was 386.78 J... which as it turned out, was not correct.
I stopped here because without knowing the MEo I could not get h.