1. Oct 22, 2007

### ~christina~

1. The problem statement, all variables and given/known data
In a jump to dunk a ball, a basketball star has a flight time of 0.852s (hang time)
His motion through space can be modeled as that of a particle at a point called his center of mass. His center of mass is at elevation of 1.02m when he leaves the floor and is at elevation 0.900m when he touches down again.

a) if he leaves the floor at an angle of 50.8 deg to the horizontal, what is the horizontal distance that he covers in the jump

b) determine the magnitude of the velocity at the instance he leaves the floor

c) what is his velocity, magnitude and direction at the instance he returns to the floor?

I drew a picture

2. Relevant equations
kinematic equations

Vf= Vi + at

Vx= Vo cos theta

Vy= Vo sin theta

Sx= Sox+ Vxt + 0.5 a t^2

3. The attempt at a solution

Well I was confused first of all about how I solve this since the initial height is higher than the final height so I guess I wouldnt' be able to just draw a line over to the other side of the trajectory to say that the initial height= 0.

Would I have to use the heights as is given?
such as the Soy= 1.85m and Sy= 0.900m?

(I just want to get this clear ..but I'll post my work after this assuming this is correct unless someone else says otherwise)

a.) to find the distance covered in the jump I think I would use the distance equation for the x component...

Sx= Sox + Vxt + 0.5axt

since a= 0 in x direction...
Vx= Vo cos theta

however for that you'd need Vo so I guess I need to plug into the y component of distanc to find the Vo...

Sy= Soy + Vyt+ 0.5at^2

Sy= 0.900m
Soy= 1.02m

Vy= Vo sin theta

a= -9.8
theta= 50.8 deg
t= 0.852s

Sy= Soy + Vyt+ 0.5at^2

0.900m= 1.85m + Vo sin (50.8) (0.852s) - 4.9 (0.852)^2

-.12= .660 Vo - 3.56

3.44 = .660 Vo

Vo= 5.21m/s

Since have Vo...plug into the x component of the distance equation

Sx= Sox + Vxt + 0.5 at^2
a= 0
Sox= 0
Sx= ?
Vx= Vo cos theta => 5.21m/s cos 50.8 = 3.29 m/s
Vo= 5.21 m/s
theta= 50.8 deg
t= 0.852s

Sx= 3.29m/s (0.852s)

Sx= 2.80 m ===> is it just me or this seems sort of small in x distance..

basically need a check on how things are going here
Thanks

Last edited: Oct 22, 2007
2. Oct 23, 2007

### learningphysics

everything looks good to me.

3. Oct 23, 2007

### ~christina~

b) determine the magnitude of his velocity at the instant he leaves the floor.

I think I found this in part a (Vo) since I had to plug that into the equation for the x distance so would it be ..

Vo=?

or
Vx= 5.21m/s cos 50.8= 3.29m/s

Vy= 5.21m/s sin 50.8= 4.04m/s

then

Vtotal= $$\sqrt{} (3.29m/s)^2 + (4.04m/s)^2$$

Vtot= 5.21m/s ==> wait..this is the same as Vo..haha.. slaps forehead* => or is it??

_____________________________________________________________

c) what is his velocity, magnitude and direction , at the instance he returns to the floor?

t= 0.852 s
Vo= 5.21m/s
Sx= 2.80m
Sy= 0.900m
Soy= 1.02m
theta= 50.8

x component stays same...
Vx= Vocos theta
Vx= 5.21cos 50.8= 3.29m/s

Vx= 3.29m/s

Vy=?
Vfy= Voy + ayt

Vfy= Vosin theta - 9.8(0.852s)
Vfy= 5.21sin 50.8 - 8.35

Vfy= -4.31m/s or just 4.31m/s ==========> is this fine? I guess even though I got a

negative I can change it to a possitive though but I'm not sure.

Vtot= $$\sqrt{} (3.29m/s)^2 + (-4.31)^2$$

Vtot= 5.21

HOW CAN THIS BE THE SAME AS BEFORE???

4. Oct 23, 2007

### Staff: Mentor

Of course it's the same.
_____________________________________________________________

Why change it? He's coming down, so Vfy should be negative. (Remember that you need to find direction as well as speed.)

Check your arithmetic on that last step.

5. Oct 23, 2007

### ~christina~

Fixed....

So vfy= -4.31m/s

and in the end I get after checking the math...It isn't the same..it's

V= 5.42m/s !

(more than the original and that makes sense sinc he's lower than the initial height thus the velocity is greater => I think that's how I'm supposed to think about that)

however for the direction...I was thinking about that..would it mean I have to find the angle??

I would think that entails the finding of the angle with tan I'm not quite sure about that...
not sure how the velocity component's look and which is which to plug into the

tan theta = y/x

would it be the velocity total that I found over the y component of the velocity?
like..so

tan theta= 5.42/ -4.31
tan theta= -1.25

theta= -51.5 ?

How is this?

Thank You Doc Al

6. Oct 23, 2007

### Staff: Mentor

Good.

Good.

No, it would be: tan theta = Vy/Vx

7. Oct 23, 2007

### ~christina~

I get is now...

so it would be

tan= Vy/Vx

tan= -4.31/ 3.29= -1.31

theta= -52.6 deg

Last edited: Oct 23, 2007