Projectile Motion of Basketball Shot to Ring

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SUMMARY

The discussion focuses on calculating the horizontal displacement of a basketball shot from a height of 2.8 meters to a ring at 3.3 meters. The basketball is released with a velocity of 5.5 m/s at an angle of 54 degrees. The initial vertical velocity is determined to be 4.45 m/s, while the horizontal velocity is calculated at 3.23 m/s. The challenge lies in determining the time of flight, as the projectile does not start and end at the same height.

PREREQUISITES
  • Understanding of projectile motion principles
  • Ability to calculate initial vertical and horizontal velocities
  • Familiarity with kinematic equations
  • Knowledge of trigonometric functions for angle calculations
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  • Calculate the time of flight using the vertical motion equations
  • Explore the use of kinematic equations for non-level projectile motion
  • Learn about the impact of initial height on projectile trajectories
  • Investigate the effects of different launch angles on horizontal displacement
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Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators looking for practical examples of kinematic equations in action.

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Homework Statement



A basketball is thrown from a height of 2.8m above the ground and goes through a basketball ring that is 3.3m above the ground. If the release velocity was 5.5 m/s at an angle of 54 degrees upwards from the horizontal, calculate the horizontal displacement the ball will travel between being released and reaching the ring

Homework Equations



I know I have to find initial vertical and horizontal velocity and time

The Attempt at a Solution



I have found initial vertical velocity of 4.45m/s and horizontal velocity to be 3.23m/s

I don't know where to start with time as i have only ever done projectiles that take-off and land at the same height ?
 
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For any vertical or horizontal displacement, the time taken are equal.
 
Last edited:

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