Using a spring gun, a 4.0-kg steel block is launched up a lubricated ramp made out of steel. The angle of the ramp was measured to be 38.8º. The inital speed of the block was 12.0 m/s.
(a) Determine the vertical height that the block reaches above its launching point.
(b) What speed does the block have when it sldes back to its startin point?
physics - Henry, Sunday, October 30, 2011 at 2:56pm
Vo = 12m/s @ 38.8 Deg.
Xo = 12cos38.8 = 9.35m/s.
Yo = 12sin38.8 = 7.5m/s.
a. h = (Yf^2 - Yo^2) / 2g,
h = (0 - (7.5)^2) / -19.6 = 2.88m.
b. Yf^2 = Yo^2 + 2g*d,
Yf^2 = 0 + 19.6*2.88 = 56.4,
Yf = 7.5m/s.