What is the ball's horizontal displacement?

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The discussion revolves around calculating the horizontal displacement of a tennis ball served horizontally at 24 m/s from a height of 2.5 m. Key questions include determining the time the ball is airborne, its horizontal displacement, velocity at impact, and the distance it clears the net. Participants suggest using kinematic equations for motion in two dimensions, emphasizing the importance of understanding the relationship between height and time of flight. The initial height of 2.5 m is relevant for calculating the time until the ball hits the ground, despite confusion about the ball's trajectory. Overall, the focus is on applying physics principles to solve the problem effectively.
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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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