Ball in Free-Fall: Find Final Speed

  • Thread starter Thread starter aaronb
  • Start date Start date
  • Tags Tags
    Ball Free-fall
Click For Summary
SUMMARY

The discussion focuses on calculating the final speed of a ball dropped from a height of 13.9 meters with an initial downward speed of 4.9 m/s, and also when thrown upwards with the same speed. For part (a), the final speed when the ball hits the table 0.7 meters below the release point is calculated to be 16.8 m/s. In part (b), when thrown upwards, the final speed upon hitting the table is determined to be 17.3 m/s. The equation used for both calculations is vf = sqrt(v(initial)^2 + 2g(h - y)).

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Familiarity with gravitational acceleration (g = 9.81 m/s²)
  • Knowledge of potential and kinetic energy concepts
  • Ability to manipulate algebraic equations for solving physics problems
NEXT STEPS
  • Study the derivation and application of kinematic equations in various scenarios
  • Explore the concepts of potential energy and kinetic energy in physics
  • Learn about the effects of air resistance on falling objects
  • Investigate the differences in motion for objects thrown upwards versus dropped
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding motion under the influence of gravity.

aaronb
Messages
22
Reaction score
0

Homework Statement



A ball of mass m = 8 kg is dropped from rest at a height h = 13.9 m above the ground. Ignore air resistance.

(a)If the ball is being released with a downward speed 4.9 m/s initially, what will be its final speed when it hits the table 0.7 m below the release point?

(b) If the ball is being thrown upwards instead with the same speed 4.9 m/s, what is the final speed when it hits the table?

Homework Equations



Kf + Ugf = Ki + Ugi

The Attempt at a Solution


vf = sqrt(v(initial)^2 + 2g(h - y))
I just plugged in numbers and got 16.8 for part (a) and I plugged in numbers for part (b) and got 17.3. Also, for part (b), I substituted y=0 because I chose the y-axis to start at the bottom where the hand would be.
 
Physics news on Phys.org
To be clear. The ball is released at 13.9m and it hits the table at .7m?

And the initial velocity is 4.9m/s.

What does dropped from rest mean when you have an initial velocity given?
 
vf = sqrt(v(initial)^2 + 2g(h - y))
In the first part the distance traveled by the ball is only 0.7m.
 
Ignore that released from rest part.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
5K
Replies
34
Views
2K
  • · Replies 100 ·
4
Replies
100
Views
6K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
14
Views
3K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K