- #1
Eclair_de_XII
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Homework Statement
"A ball of moist clay falls 15.0 m to the ground. It is in contact with the ground for 20.0 ms before stopping. (a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground? (Treat the ball as a particle.) (b) Is the average acceleration up or down?"
Homework Equations
##x_0=15m##
##x=0m##
##t_2=0.02s##
##v^2=v_0+2a(x-x_0)##
##v=v_0+at##
Answer as given by book: (a) 857 m/s2; (b) up
The Attempt at a Solution
##v^2=2(9.8\frac{m}{s^2})(15m)##
##v_0=0\frac{m}{s}##
##v=17.15\frac{m}{s}##
Next, I treat this as my next initial velocity when the clay rolls on the ground for 0.02 s.
##v_0=17.15\frac{m}{s}##
##v=0\frac{m}{s}##
##t=0.02s##
##0\frac{m}{s}=|17.15\frac{m}{s}|+a(0.02s)##
##a=|\frac{-17.15\frac{m}{s}}{0.02s}|=|-875\frac{m}{s^2}|≠857\frac{m}{s^2}##
Aside from not matching the answer in the book, the sign of the acceleration is negative, which would mean it is going down. Unless I'm mistaken, and the negative acceleration is going up from -875 m\s2 to 0 m\s2... I'm not sure if I'm correct or not, in this assumption, though. Could someone confirm or deny my answer and my assumption?