What is the ball's horizontal displacement?

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion, specifically analyzing the horizontal displacement of a tennis ball served horizontally from a certain height. The problem includes multiple parts related to the ball's time in the air, horizontal displacement, impact velocity, and clearance over a net.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss kinematic equations relevant to two-dimensional motion and question their application to the problem. There is an attempt to derive the time of flight using height, with some confusion about the reasoning behind the chosen values.

Discussion Status

The discussion is ongoing, with participants exploring various kinematic equations and their relevance to the problem. Some guidance has been provided regarding the equations needed, but there remains uncertainty about the application of these equations and the underlying concepts.

Contextual Notes

Participants express a lack of understanding regarding the initial conditions and the physical interpretation of the height from which the ball is served. There is also mention of the absence of air resistance in the problem setup.

F.B
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Ahh I am already stuck on a question. I just need help getting started.

Anyways this is the question

A tennis player serves a ball horizontally, giving it a speed of 24 m/s from a height of 2.5 m. The player is 12 m from the net. The top of the net is 0.90 m above the court surface. The ball clears the net and lands on the other side. Air resistance is negligible.
a)For how long is the ball airborne?
b)What is the horizontal displacement?
c)What is the velocity at impact?
d)By what distance does the ball clear the net.

I have no idea how to start this. I don't even know what i have. Can you please help me.
 
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Do you know any formulas for motion in 2 dimensions?
 
ya i kno all of them
 
Here are some basic kinematics equations you will need:

Vf = Vi + at
Xf = xi + ViT + 1/2aT^2
xf - xi = 1/2(vi + vf)T
Vf^2 = Vi^2 + 2a(xf - xi)

Try to look at the problem again with those in mind and see where it takes you.
 
I can get the time like this but it doesn't help because i don't understand why it works.

2.5=0 - 1/2(9.8)t^2

By why does using 2.5 for height work because doesn't the ball hit the ground at the end.
 

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