Projectile Motion and basketball dunk

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SUMMARY

The discussion focuses on calculating the time of flight for a basketball player who jumps to dunk the ball, covering a horizontal distance of 2.80m. The player’s center of mass starts at an elevation of 1.02m, reaches a maximum height of 1.85m, and lands at 0.900m. Key equations used include the kinematic equations for projectile motion, specifically for vertical and horizontal components. The total time of flight is determined by summing the time to reach maximum height and the time to descend back to the ground level.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Knowledge of projectile motion principles
  • Familiarity with gravitational acceleration (g = 9.80 m/s²)
  • Ability to manipulate algebraic equations for solving time and height
NEXT STEPS
  • Study the derivation of kinematic equations for projectile motion
  • Learn how to calculate maximum height and time of flight in projectile motion
  • Explore the effects of varying launch angles on projectile trajectories
  • Investigate real-world applications of projectile motion in sports
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Students studying physics, particularly those focusing on mechanics and projectile motion, as well as coaches and athletes interested in optimizing jumping techniques for sports like basketball.

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.
 
Last edited:
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omegax013 said:

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


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?
 
i am air headed sorry to waste anyone's time reading this.
 
Last edited:

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