Equations of Kinematics in 2-D/Projectile Motion

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SUMMARY

The discussion centers on solving a projectile motion problem involving a tennis ball struck horizontally at a speed of 28.0 m/s, which travels a horizontal distance of 19.6 m before hitting the court. The correct height of the ball when it leaves the racket is determined to be 2.40 m. Participants clarify that the maximum height of the ball coincides with its launch height, emphasizing the importance of understanding the initial conditions of projectile motion. Misinterpretations regarding the trajectory and calculations are addressed, leading to a resolution of the problem.

PREREQUISITES
  • Understanding of kinematic equations, specifically for projectile motion.
  • Familiarity with horizontal and vertical motion components in physics.
  • Knowledge of basic algebra for solving equations.
  • Experience with interpreting and applying physics problem statements.
NEXT STEPS
  • Study the derivation and application of the kinematic equations for projectile motion.
  • Learn about the effects of air resistance on projectile trajectories.
  • Explore the concept of initial velocity and its impact on projectile height and distance.
  • Practice solving various projectile motion problems with different initial conditions.
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding the principles of projectile motion in two dimensions.

Michele Nunes
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Homework Statement


A tennis ball is struck such that it leaves the racket horizontally with a speed of 28.0 m/s. The ball hits the court at a horizontal distance of 19.6 m from the racket. What is the height of the tennis ball when it leaves the racket?

Homework Equations


v = vo + at
x = 1/2(vo + v)t
x = vot + 1/2(at2)
v2 = vo2 + 2ax

The Attempt at a Solution


I tried to use symmetry in the trajectory but it just kept coming out weird and not making sense, I tried to split up the trajectory into segments, like from the initial position to the max height and then the max height back to the original horizontal position and then that position to the ground, I was trying to find the total vertical displacement for the second half of the trajectory and then the total vertical displacement for the first half and then I was planning on subtracting the two to find the difference of height which would be the answer but my calculations just weren't coming out right at all.
 
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by the way the answer is 2.40 m
 
Hey,
The ball is released horizontally so the max height of the ball is at the position it was launched. I think you just misunderstood the set up of the problem.
 
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CAF123 said:
Hey,
The ball is released horizontally so the max height of the ball is at the position it was launched. I think you just misunderstood the set up of the problem.
Ohhhhh, okay the universe has aligned, thank you so much!
 
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