Asked by Steve
With what velocity should the ball be kicked (magnitude in m/s and direction) for an ideal kick off? 4.8 second hang time and horizontal displacement of 60 yards and how high in meters should the ball be expected to go?
Thank you
Thank you
Answers
Answered by
Henry
Tr + Tf = 4.8s. = Rise time + Fall time = Time in air.
Tr = Tf = 4.8/2 = 2.4 s.
Dx = 60yds * 3Ft/yd * 1m/3.3Ft=54.55 m.
Dx = Xo * (Tr+Tf) = 54.55 m.
Xo * 4.8 = 54.55
Xo = 11.36 m/s. = Hor. component of
initial velocity.
Y = Yo + g*Tr = 0 @ max ht.
Yo - 9.8*2.4 = 0
Yo = 23.52 m/s. = Ver. component of initial velocity.
tan A = Yo/Xo = 23.52/11.36 = 2.07042
A = 64.2o
Vo=Yo/sin A=23.52/sin64.2=26.1m/s[64.2o]
= Initial velocity.
h max = (Y^2-Yo^2)/2g
h max = (0-23.52^2)/-19.6 = 28.2 m.
Tr = Tf = 4.8/2 = 2.4 s.
Dx = 60yds * 3Ft/yd * 1m/3.3Ft=54.55 m.
Dx = Xo * (Tr+Tf) = 54.55 m.
Xo * 4.8 = 54.55
Xo = 11.36 m/s. = Hor. component of
initial velocity.
Y = Yo + g*Tr = 0 @ max ht.
Yo - 9.8*2.4 = 0
Yo = 23.52 m/s. = Ver. component of initial velocity.
tan A = Yo/Xo = 23.52/11.36 = 2.07042
A = 64.2o
Vo=Yo/sin A=23.52/sin64.2=26.1m/s[64.2o]
= Initial velocity.
h max = (Y^2-Yo^2)/2g
h max = (0-23.52^2)/-19.6 = 28.2 m.
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