Gravitational potential energy of baseball

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Homework Help Overview

The problem involves a baseball thrown from a height, requiring the application of energy conservation principles to determine its speed just before impact with the ground. The scenario includes two parts with different launch angles and a consideration of air resistance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the transformation of potential energy to kinetic energy and question the necessity of mass in the calculations. There is also exploration of the initial kinetic energy due to the baseball's launch speed.

Discussion Status

Participants are actively questioning assumptions about energy types involved and whether mass is needed for the calculations. Some guidance has been provided regarding the role of mass in the energy equations, but no consensus has been reached on the final approach to finding the speed.

Contextual Notes

There is a noted absence of mass as a variable in the problem, leading to confusion among participants about its relevance. The discussion also includes the effects of air resistance on the outcomes of the different launch angles.

aftermatth
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Homework Statement


A baseball is thrown from the roof of a 22.0-m-tall building with an inital velocty of magnitude 12.0 m/s and directed at an angle of 53.1 degrees above the horizontal.
a) What is the speed o fthe ball just before it strikes the ground? Use energy conservation and ignore air resistance.
b) What is the answer for part (a) if the inital velocity is at an angle of 53.1 below the horizontal?
c) If the effects of air resistance are included, will part (a) or part (b) give the higher speed?

Vo = 12 m/s
y1 = 22m
y2 = 0m
a = -9.8 m/s²
Voy=(12m/s)sin53.1 = 9.6 m/s
don't know why they don't give any variables for mass of the baseball.. ;\
other problems use .145kg for baseball so i decided just to use that

Homework Equations


E= K + Ugrav = constant
K= 1/2mv1²
1/2mv1² + mgy1 = 1/2mv2² + mgy2

The Attempt at a Solution


Initially, I'm guessing that the baseball only has potential energy, and when it hits the ground it's all transformed into kinetic energy.

K1 + Ugrav1 = K2 + Ugrav2
0 + Ugrav1 = K2 + 0
Ugrav1 = K2 (initial gravitational potential engergy = final kinetic energy)
K2 = 1/2mv2² = (1/2)(.145kg)(9.6m/s)² = 6.7J
Ugrav1 = mgy1 = (.145kg)(9.8m/s²)(22m) = 31.2 J
and I'm stuck.

how do i find the speed?

for part (b) i know that the velocity will be the same as the first part, because when it returns back from throw point, it will end up having the same velocity as if it were thrown 53.1 degrees above.
 
Last edited:
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aftermatth said:
Initially, I'm guessing that the baseball only has potential energy, and when it hits the ground it's all transformed into kinetic energy.
You are given the initial speed of the ball, so it has kinetic energy.
 
Incidentally, the mass of the ball is NOT needed to solve this problem.

Initially, I'm guessing that the baseball only has potential energy, and when it hits the ground it's all transformed into kinetic energy.

Does the ball only have potential energy to start with? Isn't it launched at a certain speed?
 
Oops, sorry for repeating what Doc Al said
 
Doc Al said:
You are given the initial speed of the ball, so it has kinetic energy.

oh youre right it does have kinetic energy..
K1 + Ugrav1 = K2

mass isn't required? do i just omit it from all parts?
i still don't know how to get to the velocity.
 
Last edited:
You will find if you divide both sides of:
K1 + Ugrav1 = K2
by the mass, the problem does not contain mass.
 

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