What is the ball's maximum height?

Click For Summary

Homework Help Overview

The problem involves a ball thrown upward, with the goal of determining its maximum height given an initial vertical speed of 10.8 m/s and an acceleration due to gravity of 9.8 m/s², while neglecting air resistance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to relate initial speed, final speed, acceleration, and distance. There are questions regarding the definitions of variables used in the equations, particularly concerning the signs of speeds and the meaning of terms like 'u' and 'is'.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the kinematic equations. Some guidance has been offered regarding the relationships between the variables, but there is no explicit consensus on the approach to finding the maximum height.

Contextual Notes

There are indications of confusion regarding the signs of the velocities and the assignment of values to variables in the equations, which may affect the calculations. The original poster has reiterated the problem statement and equations, suggesting a need for clarification on the setup.

Casey314stl
Messages
15
Reaction score
0

Homework Statement


A ball is thrown upward.
What is its maximum height? Its initial
vertical speed is 10.8 m/s and the acceleration
of gravity is 9.8 m/s^2
Neglect air resistance.

Answer in units of m

Homework Equations



dfinal=dintial+vfinalΔt+ 1/2averageaccelerationΔt^2 or v^2final=v^2initial +2averageaccelerationΔd

The Attempt at a Solution


14.2463
 
Last edited:
Physics news on Phys.org
No remember that since the object is traveling up it will have a negative speed therefore:

Vf^2=u^2+2is
 
mtayab1994 said:
No remember that since the object is traveling up it will have a negative speed therefore:

Vf^2=u^2+2is
what does u and is stand for?
 
u and v are vectors v is the final speed which in your case will be 0 because the kinetic energy will be 0 at the maximum height so you will be left with 0=speed of the ball -2(speed of gravity)s.
 
so how do i use this to find the height?
 
0=(initial speed)^2-2(speed of gravity)s then once you have this down just solve the equation.
 
Last edited:
mtayab1994 said:
0=(initial speed)^2-2(speed of gravity)s than once you have this down just solve the equation.
I am solving for Vf^2 which is vertical speed final?
so I plug in 0-2(10.8)=-21.6?
 
Casey314stl said:

Homework Statement


A ball is thrown upward.
What is its maximum height? Its initial
vertical speed is 10.8 m/s and the acceleration
of gravity is 9.8 m/s^2
Neglect air resistance.

Answer in units of m


Homework Equations



dfinal=dintial+vfinalΔt+ 1/2averageaccelerationΔt^2 or v^2final=v^2initial +2averageaccelerationΔd

The Attempt at a Solution


14.2463
Your second equation looks promising. In more compact terms:

vf2 = vi2 + 2*a*d

where vi is the initial velocity, vf the final velocity, a is the acceleration and d is the distance. The distance d is what you're looking for. What values will you assign to the other variables?
 

Similar threads

Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
11
Views
2K
Replies
7
Views
3K
Replies
7
Views
3K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 15 ·
Replies
15
Views
3K