What Is the Speed of a Cannonball at Its Maximum Height?

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Homework Help Overview

The discussion revolves around a physics problem concerning the speed of a cannonball at its maximum height, specifically in the context of projectile motion. The original poster presents a scenario where a cannonball is launched and reaches a maximum height, seeking to determine its speed at that point.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply various physics equations to find the speed of the cannonball at maximum height, including gravitational force and kinematic equations. Some participants question the assumptions made regarding the information provided, particularly the lack of initial velocity or angle of launch.

Discussion Status

The discussion is ongoing, with participants noting the insufficiency of the given information to solve the problem. There is an acknowledgment that additional details, such as the initial speed or launch angle, are necessary for a complete analysis. The original poster plans to seek clarification from their teacher.

Contextual Notes

Participants highlight that the maximum height corresponds to a vertical velocity of zero, indicating that only horizontal velocity remains, which complicates the ability to solve the problem with the current data.

Chaalie
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Hey guys!
This is my first post on this forum, though I've been looking through it for a few months now. I am in a basic Grade 11 physics course, and I'm having a little trouble with this one. So, here goes...

Homework Statement


A cannonball (mass=15.5kg) is shot out of a cannon on Earth (assume it is shot at a height of 0m). The maximum height that it reaches is 20.1m. What is the speed of the cannonball at the maximum height?

Homework Equations


The different equations I know that might be useful are...
Eg=m*g*h
V=d/t
A=v/t
d=V(initial)*t + (A*(t)^2)/2
d=[(V(initial)+V(final))*t]/2

The Attempt at a Solution


My attempts have been a little rough, but here is my best one...
F=mg
=15.5*9.8
=151.9N

A=F/m
=151.9N/15.5
=9.8m/s^2

Vf^2=Vi^2 + 2A*d
= 0 + 2(9.8)(20.1)
Vf^2= 393.96
Therefore, Vf = 19.84842563m/sI have a feeling I've done something wrong...any input?
I'd still like to figure it out myself, so if somebody could just point me in the right direction, that'd be great!Charlie
 
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The the maximum height, its vertical velocity is zero. It has only horizontal velocity, and horizontal velocity depends on the velocity at which it was fired. It isn't possible to solve this problem with the information you are given.
 
cavalier said:
It isn't possible to solve this problem with the information you are given.

Correct. We need to know the initial speed or angle, or the initial horizontal component of velocity, to solve the problem.
 
Thanks guys!
I'll check with my teacher to see if he missed anything.
 

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