Constant acceleration ball throw problem

In summary: Now you should do the problem with the other equation, the one you used first. It should give the same answer.Much better. Now you should do the problem with the other equation, the one you used first. It should give the same answer.In summary, the entertainer needs to throw the ball upward with a minimum initial speed of 23.7 m/s and the ball reaches a height of 28.7 meters above its initial position just as she reaches the table. The equations used were x(t) = vt - (1/2)at^2 + x0 and x = x0+v0t+(1/2)at2.
  • #1
Crusaderking1
159
0

Homework Statement



An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in the air, she runs to and from a table 5.80 m away at a constant speed of 2.40 m/s, returning just in time to catch the falling ball.

1. With what minimum initial speed must she throw the ball upward to accomplish this feat?

I'm positive the answer is 23.7 m/s.

2. How high above its initial position is the ball just as she reaches the table?

Homework Equations



A constant acceleration equation.

The Attempt at a Solution



After I obtained the speed, I simply plugged it into the equation -9.8(1.2085)2+23.7(1.2085)

and received 14.32 m for problem number 2. Is this right? I don't feel like it is. Thanks!
 
Last edited:
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  • #2
Crusaderking1 said:
After I obtained the speed, I simply plugged it into the equation -9.8(1.2085)2+23.7(1.2085)

and received 14.32 m for problem number 2. Is this right? I don't feel like it is. Thanks!
How did you get 1.2085 s?
 
  • #3
kuruman said:
How did you get 1.2085 s?

Split the time in half.

5.80/2.40 = 2.417, and then halved it.
 
  • #4
You halved one time too many. You forgot that she runs a total distance of 5.8+5.8 =11.6 m. So the total running time is 11.6/2.4 not 5.8/2.4.
 
  • #5
kuruman said:
You halved one time too many. You forgot that she runs a total distance of 5.8+5.8 =11.6 m. So the total running time is 11.6/2.4 not 5.8/2.4.

ooh ok thanks! So I need to plug 2.417 instead of 1.2085 into the equation.
 
  • #6
I got 33.6, sound better?
 
  • #7
Am I using the wrong equation now? I received wrong answer for using 2.417 in the equation above.

x(t) = vt - (1/2)at^2 + x0 sounds better.

23.7(2.417)-0.5(-9.8)(2.417)^2 + 0?

xo = 0?
 
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  • #8
Crusaderking1 said:
Am I using the wrong equation now? I received wrong answer for using 2.417 in the equation above.

x(t) = vt - (1/2)at^2 + x0 sounds better.

23.7(2.417)-0.5(-9.8)(2.417)^2 + 0?

xo = 0?
You have one minus sign too many. In the equation x = x0+v0t+(1/2)at2 you should replace a with -9.8 m/s2.

Yes x0 = 0.
 
  • #9
kuruman said:
You have one minus sign too many. In the equation x = x0+v0t+(1/2)at2 you should replace a with -9.8 m/s2.

Yes x0 = 0.

Ok thanks a ton! I received 28.7 meters. sound better?
 
  • #10
Much better.
 

FAQ: Constant acceleration ball throw problem

What is the constant acceleration ball throw problem?

The constant acceleration ball throw problem is a physics problem that involves calculating the trajectory of an object (in this case, a ball) thrown with a constant acceleration, such as the force of gravity. It is used to understand the motion of objects under the influence of a constant force.

How do you calculate the velocity and position of the ball in the constant acceleration ball throw problem?

To calculate the velocity and position of the ball, you can use the equations of motion, specifically the equations for constant acceleration. These equations include the initial velocity, acceleration, and time, and can be used to determine the final velocity and position of the ball at any given time.

3. What is the difference between constant acceleration and constant velocity in the ball throw problem?

Constant acceleration refers to the rate at which the velocity of an object changes over time, while constant velocity means the object is moving at a constant rate without any changes in its speed or direction. In the ball throw problem, constant acceleration is caused by a constant force, such as gravity, while constant velocity means the ball is not experiencing any net force.

4. How does air resistance affect the constant acceleration ball throw problem?

Air resistance can significantly affect the motion of the ball in the constant acceleration ball throw problem. As the ball travels through the air, it experiences a force in the opposite direction of its motion, which can slow it down and change its trajectory. This means that the equations of motion used to calculate the velocity and position of the ball may need to be adjusted to account for air resistance.

5. What real-life applications does the constant acceleration ball throw problem have?

The constant acceleration ball throw problem has many real-life applications, such as understanding the motion of projectiles like a baseball being thrown or a rocket launching into space. It is also used in sports, such as calculating the trajectory of a golf ball or basketball. In addition, it is used in engineering and design to predict the motion of objects under the influence of various forces.

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