How Does Friction Affect the Ball's Velocity at the Table Edge?

[cheerspens]

Homework Statement

The ball is moving at 2.0m/s just 40cm before it hits the table edge. If the ball has a mass of 51.3 grams and experiences a 0.23 N frictional force between it and the table, how fast will it be moving when it reaches the table's edge?

The table is 0.92m tall. How long will it take the ball to hit the ground?

Homework Equations

X=X_o+V_ot+(0.5)at²
Y=Y_o+V_oyt+(0.5)at²
X=X_o+V^oxt
F_NET=ma
F_g=mg
Quadratic Formula

The Attempt at a Solution

I used F_NET=ma to solve for the acceleration and got -4.48m/s². Then I used X=X_o+V_ot+(0.5)at² and the quadratic formula to solve for t and got 0.3 sec. I then plugged this into V=V_o+at to get a velocity of 0.656m/s. Is this correct?

For the second question: How long will it take for the ball to hit the ground?
I seemed to be stumped. Maybe my velocity I solved for is incorrect or I'm not even setting it up right?

Thanks in advance for any help.

 

[ehild]

Your result for the horizontal velocity on the table is correct.

When the ball is falling from the table, its motion can be decomposed into a horizontal motion with constant velocity (what you just calculated) and a vertical motion (free fall) with constant acceleration g. The time of reaching the ground is independent of the horizontal velocity.

ehild

 

[cheerspens]

So the answer I'm getting is that it will take 0.09 sec for the ball to hit the ground and it will land 0.1719 meters away from the table.
Does this seem correct?

 

[ehild]

No. How did you get 0.09 s?

ehild

 

[cheerspens]

So if I were to set up a variable list for how long it will take the ball to hit the ground would it look like the following?

X_o=0m
X=?
V_ox=1.91m/s
t=?
Y_o=0.92m
Y=0m
V_oy=-9.8m/s
g=-9.8m/s²

Using these values I got 0.09 seconds.

 

[ehild]

I do not see how did you get Voy=-9.8m/s:

ehild

 

[cheerspens]

o I have redone this problem making some corrections and have come up with these answers:

-When the ball reaches the table's edge it will be moving 0.65m/s.
-It will take the ball 0.44 seconds to hit the ground.
-The ball will land 0.286 meters away from the table?

Have I gotten this problem correct?

And if I were to graph my y-component velocity, would it be a constant 0m/s?

Thanks!

 

[ehild]

Your calculation is correct now.

vy is the vertical velocity. Do you really mean that it is constant and equal to zero?

ehild

 

[cheerspens]

Because V_oy is 0m/s I'm not sure how to graph the v_y component. Once the ball hits the ground doesn't that also mean the velocity is 0m/s there?

 

[ehild]

Because V_oy is 0m/s I'm not sure how to graph the v_y component. Once the ball hits the ground doesn't that also mean the velocity is 0m/s there?

No, falling objects do hit the ground with some velocity. Their velocity becomes zero after they touch the ground, because of the interaction with it.

Zero velocity means no motion. Do you really think that a ball which falls of the table , will not move downward? How does it get to the ground?

ehild

 

[cheerspens]

How do you solve for a vertical V_f?
Would it be like solving the equation V=[tex]\Delta[/tex]X/[tex]\Delta[/tex]t?

 

[ehild]

It is average velocity, but let's start with it. What is the average vertical velocity of the ball?

 

[cheerspens]

2.091m/s

 

[ehild]

2.091m/s

You see, the vertical (y) velocity component can not be constant as it is zero at the beginning of the fall and about 2.1 in m/s in average. It increases with time, the ball accelerates. What is the magnitude of this acceleration?

ehild

 

[Champdx]

f=ma
0.23N=0.0513a
a=4.48m/s^2

v^2=u^2+2as
v^2=(2)^2+2(4.48)(40)
v^2=360
v=19m/s

s=ut+1/2at^2
0.92=(19)t+1/2(9.8)t^2
4.9t^2+19t-0.92=0
t=0.05s

Correct me if I'm wrong
Thanks

 

[ehild]

What force acts on the ball after it left the table and moves in air?