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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=0.440s

Then 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=0.412s

t1+t2=t

0.440s+0.412s=0.852

Which seems too short, but maybe not.

Any help is appreciated.

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