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Kinematics of the human canon fly zaccini

  • Thread starter Newton86
  • Start date
59
0
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 ?
 

Answers and Replies

59
0
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 ?
 
25
0
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
 
59
0
Thanks for answers
butI dont 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
 
25
0
Thanks for answers
butI dont 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
 
59
0
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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