Position Velocity Acceleration vectors - Two-dimensional motion

Problem Statement:

A tennis player throws the ball against a vertical wall 25 m away (we will call this distance d). The ball is initially 2 m above the ground (y0 = 2 m), and has an initial velocity given by: v0 = 20 i+10 j (m/s). The wind produces a constant horizontal acceleration a = -3 i (m/s2).

  1. Convert the initial velocity vector from component form into magnitude and direction form.
  2. Calculate the position, velocity and acceleration vectors as a function of time.
  3. The time at which the ball reaches the highest point of its trajectory.
  4. The time at which the ball hits the wall.
  5. The height at which the ball hits the wall
  6. The velocity vector at the moment of impact.

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Solution:

The following figure shows the Cartesian axes that we are going to use to solve the problem, as well as the initial position, initial velocity and acceleration of the ball (they are not drawn to scale).

The magnitude of the initial velocity (ie the initial speed) is given by:

To determine the direction of the vector we are going to place the Cartesian axes at the origin of the velocity vector. Then, using a bit of trigonometry, we can determine the angle α from its components vx and vy.


The acceleration vector is the vector sum of the horizontal acceleration produced by the wind and the acceleration due to gravity. We will round the magnitude of the last one to g = 10 m/s2.

With the sign criterion for the Cartesian axes represented in the first figure, the acceleration vector is given by:


Now that we have calculated the acceleration vector we can integrate it to calculate the velocity. If you want to see in more detail how this integral is done, please check Problem 1.

Subtituting v0 and simplifying:

We can write separately each component of the velocity vector:


We will follow a similar procedure to calculate the position vector, but now we have to integrate the velocity vector we have just calculated.

The initial position vector is r0 = 2 j, because the ball is initially 2 m above the ground.

Integrating:

Subtituting the initial position vector and simplifying:

The components of the position vector are:


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The ball reaches the highest point of its trajectory when the vertical component of its velocity (which is given by equation (2)) equals to zero:

To hit the wall the ball has to travel the distance d. Therefore the horizontal component of its position vector (given by equation (3)) equals to d:


To calculate the height at which the ball hits the wall we substitute this time into equation (4):

We find the velocity vector by substituting the same time in equations (1) and (2), or directly in the expression of the velocity vector.

In the figure below a plot of the ball’s trajectory is shown. As you can see, it is not a parabola. A projectile has a parabolic trajectory when its acceleration is only due to gravity.

Do not forget to include the units in the results.In this problem we have rounded the acceleration of gravity to g = 10 m/s2.

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