Equations of Kinematics in 2-D/Projectile Motion

In summary, the problem involves a tennis ball being struck horizontally with a speed of 28.0 m/s and hitting the court at a horizontal distance of 19.6 m from the racket. The task is to find the height of the ball when it leaves the racket. Various equations were attempted, such as using symmetry in the trajectory and splitting up the trajectory into segments, but the correct approach was to recognize that the ball was released horizontally, making its maximum height at the position it was launched. After this realization, the answer was found to be 2.40 m.
  • #1
Michele Nunes
42
2

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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  • #2
by the way the answer is 2.40 m
 
  • #3
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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  • #4
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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FAQ: Equations of Kinematics in 2-D/Projectile Motion

1. What is the difference between 1-D and 2-D kinematics?

1-D kinematics deals with motion in a single direction, while 2-D kinematics deals with motion in two perpendicular directions simultaneously.

2. What are the basic equations of 2-D kinematics?

The basic equations of 2-D kinematics are displacement equations (x and y), velocity equations (vx and vy), and acceleration equation (a).

3. How do you solve for the initial velocity in projectile motion?

To solve for the initial velocity in projectile motion, you can use the equation v0 = v*cos(theta), where v is the initial velocity, and theta is the angle of launch.

4. What are the key concepts in projectile motion?

The key concepts in projectile motion include the independence of vertical and horizontal motion, the parabolic path of a projectile, and the constant acceleration due to gravity.

5. How do you calculate the maximum height and range of a projectile?

The maximum height of a projectile can be calculated using the equation h = (v*sin(theta))^2 / 2g, where v is the initial velocity and theta is the angle of launch. The range can be calculated using the equation R = (v^2*sin(2*theta)) / g.

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