A 0.34 kg, 0.22 m diameter ball is dropped from rest from the origin at yo = 0.
ID: 1687051 • Letter: A
Question
A 0.34 kg, 0.22 m diameter ball is dropped from rest from the origin at yo = 0.a) What is the ball's position at time t = 3.4 s?
b) What is the ball's velocity at time t = 3.4 s?
c) How long did it take the ball to fall half the distance calculated in part (a) above?
d) What was the ball's velocity when it was at half the distance calculated in part (a) above?
e) How long did it take the ball to get to half the velocity calculated in part (b) above?
f) What was the ball's position when it had half the velocity calcualted in part (b) above?
Explanation / Answer
solution: (a) from the equation of kinematics s = ut+1/2at^2 initial velocity u = 0 hence s = 1/2*9.8*(3.4)^2 = 56.64 m The position of the ball is 56.64 m on the negative y axis direction (b) velocity v = u+at = 9.8*3.4 = 35.72 m/s (c) displacement s = ut+1/2at^2 28.32 = 1/2*9.8*t^2 t^2 = 28.32/4.9 = 5.77 t = 2.44 sec (d) d = 28.32 m t = 2.44 sec v = u+at = 0+9.8*2.44 = 23.9 m/s (e) v = u+at 17.86 = 9.8*t t = 17.86/9.8 = 1.8sec (f) v = 17.86 m/s t = 1.8 sec d = v*t = 17.86*1.8 =32.14 m
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