# Projectile Motion and basketball dunk

1. Sep 19, 2008

### omegax013

1. The problem statement, all variables and given/known data

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

2. Relevant 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

3. 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.

Last edited: Sep 20, 2008
2. Sep 19, 2008

### LowlyPion

Welcome to PF.

You've got more than you may realize.

For instance you have the change in height of the center of mass to the max height. Which can yield you the time to max height by 1/2*g*t2

Likewise you have the time to fall again with the same equation but using the difference in height going down.

Doesn't adding those 2 times together give you the total time?

3. Dec 15, 2008

### DEMJ

i am air headed sorry to waste anyone's time reading this.

Last edited: Dec 15, 2008