Posted by Elaine on Wednesday, September 19, 2012 at 8:51am.
vertical motion
v = 9.81 t
y = 2 - .5(9.81)t^2
horizontal motion
v = u which is constant
x = u t
parabola
so t = x/u
y = 2 - 4.9 x^2/u^2
now the ball.
You did not tell me where the center of the ball is on the floor. I will assume it is at x = .85 and y = .85
If we just want to clear the top of the ball then the trajectory passes through the point (.85,1.7)
1.7 = 2 - 4.9 (.85)^2/u^2
-.3 = - 3.54/u^2
u = 3.44 m/s
Of course I should check if the Bob hits the ball further down on the far side after clearing the top. You can do that by solving for any other intersection of Bob and ball at x>.85
Hi Damon, thank you for your answer. With u=3.44m/s, Bob will hit the ball on the far side. The instructor showed a graph of the trajectory in class. So we know u is greater than 3.44 m/s but smaller than 5 m/s (he did a trajectory graph for u=5 m/s and Bob passed the ball by a distance). What he wants is an exact u that will help Bob not hit the ball.
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