How Fast Was the Steel Ball Rolling?

  • Thread starter Thread starter halo9909
  • Start date Start date
  • Tags Tags
    Physics
Click For Summary
SUMMARY

The problem involves calculating the horizontal velocity of a steel ball rolling off a table 1.063 meters high, landing 0.229 meters away. The time of flight was determined to be 0.4657 seconds using the equation d = 0.5at², where a is the acceleration due to gravity (9.8 m/s²). The horizontal velocity was then calculated using the formula vₓ = Δx/t, resulting in a final velocity of 0.4916 m/s. The initial calculation was confirmed correct after resolving a calculator error.

PREREQUISITES
  • Understanding of basic kinematics, specifically projectile motion
  • Familiarity with the equations of motion, particularly d = 0.5at² and vₓ = Δx/t
  • Knowledge of gravitational acceleration (9.8 m/s²)
  • Basic proficiency in using calculators for physics problems
NEXT STEPS
  • Study projectile motion in two dimensions, focusing on horizontal and vertical components
  • Learn more about the effects of air resistance on projectile motion
  • Explore advanced kinematic equations for varying acceleration scenarios
  • Practice similar problems involving different heights and distances to solidify understanding
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and kinematics, as well as educators looking for practical examples of projectile motion calculations.

halo9909
Messages
37
Reaction score
0

Homework Statement



A steel ball rolls with a constant velocity across a table top 1.063 m high. It rolls off and hits the ground +0.229 m horizontally from the edge of the table. How fast was the ball rolling?


Homework Equations



d=.5at^2
X=Vxt

The Attempt at a Solution



I first did
1.063=.5(9.8)t^2
got the time to be .4657 seconds
then put it in the X=Vxt
and got Vx to be .4916, and I am not sure where to go from there
 
Physics news on Phys.org


You approached the question correctly, but your final result is wrong.

Using [tex]\Delta x=v_{x}t[/tex] where:
[tex]\Delta x[/tex]=displacement
[tex]v_{x}[/tex]=horizontal velocity
[tex]t[/tex]=time

the displacement will be the distance it traveled from the edge of the table, the time is what you just calculated, and all that is left is the horizontal velocity.

Rearrange the formula to make velocity the subject: [tex]v_{x}=\frac{\Delta x}{t}[/tex] and substitute to find your answer.

EDIT: Sorry your answer was correct, calculator error on my part.
That is the answer to the question, in metres/second (ms-1)
 
Last edited:

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
14
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
5K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
5K