How long does it take for a ball to hit the ground from a table edge?

In summary: It is a typo.-The ball will land 0.286 meters away from the table?Correct.-And if I were to graph my y-component velocity, would it be a constant 0m/s?Yes, since the initial and final velocities in the y-direction are both 0 m/s, the velocity will remain constant at 0 m/s throughout the ball's motion.
  • #1
cheerspens
92
0

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=Xo+Vot+(0.5)at2
Y=Yo+Voyt+(0.5)at2
X=Xo+Voxt
FNET=ma
Fg=mg
Quadratic Formula

The Attempt at a Solution



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

For the second question: I have 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?

Thanks in advance for any help.
 
Last edited:
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  • #2
In the first part, if you use
vf^2 = vi^2 - 2*a*s, you get slightly different vf value.
In the second part, your t value is wrong. In that case the initial velocity in the vertical direction is zero.
 
  • #3
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?

Xo=0m
X=?
Vox=1.91m/s
t=?
Yo=0.92m
Y=0m
Voy=-9.8m/s
g=-9.8m/s2

Using these values I got 0.09 seconds. But that does not seem right...
 
  • #4
Yo=0.92m
Y=0m
Voy=-9.8m/s This is not correct. It should be zero.
g=-9.8m/s2
 
  • #5
So 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!
 
Last edited:
  • #6
-It will take the ball 0.44 seconds to hit the ground.
Time t is 0.43 s.
 

1. What are vectors and how are they used in velocity problems?

Vectors are mathematical quantities that have both magnitude (size) and direction. In velocity problems, vectors are used to represent the speed and direction of an object's motion.

2. How do you calculate the magnitude of a vector?

The magnitude of a vector can be calculated using the Pythagorean theorem, which states that the magnitude is equal to the square root of the sum of the squares of its components. In other words, to find the magnitude, you square the x and y components of the vector, add them together, and then take the square root of the result.

3. What is the difference between velocity and speed?

Velocity and speed are both measures of an object's motion, but they have different meanings. Velocity is a vector quantity that describes an object's speed and direction, while speed is a scalar quantity that only describes an object's rate of motion.

4. How do you add or subtract vectors?

To add or subtract vectors, you must first break them down into their x and y components. Then, you can simply add or subtract the corresponding components to get the resultant vector. To add vectors, you can also use the parallelogram method or the head-to-tail method.

5. Can the magnitude of a vector be negative?

Yes, the magnitude of a vector can be negative. The magnitude of a vector only describes its size, so it can be positive or negative depending on its direction. However, in most cases, we use the absolute value of the magnitude to avoid negative values.

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