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omegax013
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Homework Statement
A basketball star covers 2.80m horizontally in a jump to dunk the ball. His motion through space can be modeled as that of a particle at his center of mass. His center of mass is at elevation 1.02m when he leaves the floor. It reaches a maximum height of 1.85m above the floor and is at an elevation of 0.900m when he touches down again. Determine his time of flight.
So I am using:
Yi = 1.02m
Ymax = 1.85m
Yf= 0.900m
R= 2.80m
Homework Equations
Vx=Vosin(theta)
Vy=Vocos(theta)
x=Vocos(theta)t
y=Vosin(theta)t-(1/2)gt^2
R=(Vo^2 sin(2theta))/g
Rmax=(Vo^2)/g
t=(2Vosin(theta))/g
Ymax=(Vo^2sin^2(theta))/(2g)
Yf=Yi + Vyo*t - (1/2)gt^2
The Attempt at a Solution
Not being given Vo or theta I am having trouble finding an equation that works.Here is what I am thinking:
Start at Ymax since Vyo=0 and go to Yf
Yf=Yi + Vyo*t - (1/2)gt^2
0.90m=1.85m-(1/2)gt^2
-0.95m=-(1/2)gt^2
1.90m=gt^2
t=sqrt(1.90m/9.80(m/s^2))
t1=0.440sThen start at Ymax and go to Yi
Yf=Yi + Vyo*t - (1/2)gt^2
1.02m=1.85m-(1/2)gt^2
-0.83m=-(1/2)gt^2
1.66m=gt^2
t=sqrt(1.66m/9.80(m/s^2))
t2=0.412st1+t2=t
0.440s+0.412s=0.852
Which seems too short, but maybe not.
Any help is appreciated.
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