Ball in Free-Fall: Find Final Speed

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Homework Help Overview

The problem involves a ball of mass 8 kg being dropped from a height of 13.9 m, with two scenarios: one where it is released with an initial downward speed of 4.9 m/s and another where it is thrown upwards with the same speed. The task is to find the final speed of the ball when it hits a table located 0.7 m below the release point, while ignoring air resistance.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the kinematic equation for final speed and question the meaning of "dropped from rest" given the initial velocity provided. There is also clarification on the distance the ball travels in the first part of the problem.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the initial conditions and the implications of the problem setup. Some guidance has been offered regarding the distance traveled by the ball, but no consensus has been reached on the interpretation of the initial conditions.

Contextual Notes

There is some confusion regarding the terminology used in the problem statement, particularly the phrase "dropped from rest" in the context of having an initial velocity. This may affect how participants approach the calculations.

aaronb
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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.
 
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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.
 

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