Kinematics of the human canon fly zaccini

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Homework Help Overview

The discussion revolves around a kinematics problem involving projectile motion, specifically analyzing the flight of a human cannonball. The original poster presents a scenario where the cannonball travels a distance of 53 meters with an initial velocity of 24.5 m/s at an angle with the horizontal plane, seeking to determine the launch angle and the maximum height achieved during the flight.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to relate distance, velocity, and angle. There are attempts to derive the angle using the relationship between horizontal and vertical components of motion. Questions arise regarding the correct formula and values to use, particularly concerning the initial velocity.

Discussion Status

Some participants have provided formulas and attempted to derive the angle, leading to different results. There is acknowledgment of a potential error in the initial velocity value, which has prompted further exploration of the calculations. While some participants express confusion, others indicate they have found clarity through shared resources.

Contextual Notes

There is a noted discrepancy in the initial velocity value used in calculations, which has influenced the results. Participants are also navigating the implications of using different trigonometric functions in their approaches.

Newton86
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as a human canon fly zaccini flied a distance of 53 meter in 1940 ( still record )
his velocity as he leave the cannon is 24,5 m/s in a angle a with thw horizontalplane

a) what is the angle ? ( 31,2 )
b) find the highest point during the record ( 8,04m)

can you set up theese tasks ?
 
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Newton86 said:
as a human canon fly zaccini flied a distance of 53 meter in 1940 ( still record )
his velocity as he leave the cannon is 24,5 m/s in a angle a with thw horizontalplane

a) what is the angle ? ( 31,2 )
b) find the highest point during the record ( 8,04m)

can you set up theese tasks ?

Do anyone know the formula ?
 
x = v0*cos(a)*t
y = v0*sin(a)*t - 1/2*gt^2
since when x = 53 m, y = 0
v0*sin(a)*t - 1/2*gt^2 = 0
so t = 2v0*sin(a)/g
then x = v0*cos(a)*2v0*sin(a)/g = v0^2*sin(2a)/g
sin(2a) = g*x/v0^2
plug known values and get a = 30 degrees (not 31.2 (?))
then we can find t.
highest: find y when t' = t/2
 
Thanks for answers
butI don't get it :(
g*x/vo^2 I get 0,88 ? ( 9,8*53/24,2^2)

And the reason for 30 degress was that I did put in wrong velocity It should be 24,2m/s
 
Newton86 said:
Thanks for answers
butI don't get it :(
g*x/vo^2 I get 0,88 ? ( 9,8*53/24,2^2)

And the reason for 30 degress was that I did put in wrong velocity It should be 24,2m/s

yes, but don't put 0.88 in your calculator. Just use calculator for 1/2 arcsin(g*x/vo^2) = 31.2
 
Yes Thanks =) I got it now=) Found a formula on the link kamerling showed, I did not use arcsin just sin.
 
Last edited:

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