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November 28, 2014

November 28, 2014

Posted by **Elaine** on Wednesday, September 19, 2012 at 8:51am.

If we use kinematics equations for projectile, the answer is 2.66 m/s; however, with this velocity the person will land on the ball. I need to find the velocity at which the person will just miss the ball. And I know this velocity is less than 5 m/s. Thank you for your help.

- Physics -
**Damon**, Wednesday, September 19, 2012 at 9:20amvertical 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

- Physics -
**Elaine**, Wednesday, September 19, 2012 at 9:51amHi 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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