# Equations of Kinematics in 2-D/Projectile Motion

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1. Sep 27, 2015

### Michele Nunes

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
v = vo + at
x = 1/2(vo + v)t
x = vot + 1/2(at2)
v2 = vo2 + 2ax
3. 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.

2. Sep 27, 2015

### Michele Nunes

by the way the answer is 2.40 m

3. Sep 27, 2015

### CAF123

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.

4. Sep 27, 2015

### Michele Nunes

Ohhhhh, okay the universe has aligned, thank you so much!

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