Position Velocity Acceleration vectors - Parabolic and uniform motions

Problem Statement:

A boy throws a ball from the origin of coordinates (see figure). The ball has an initial velocity v0 (m/s) whose direction is α = 600. The boy wants the ball to fall into the back of a truck (whose length is L = 15 m) moving at a constant velocity vC = 20 i (m/s). The initial distance between the truck and the boy is d0 = 10 m.

Calculate:

  1. The minimum value of v0 for the ball to fall into the back of the truck.
  2. The maximum value of v0 for the ball to fall into the back of the truck.

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Solution:

First of all we are going to write the position vector of the ball as a function of time. In order to do that, we integrate the acceleration vector to obtain the velocity and then integrate the velocity vector to obtain r. The acceleration of the ball is due to gravity:

The initial velocity vector of the ball in terms of its components is:

If you want to see in more detail how these components are calculated check out Problem 6.

We obtain the velocity vector as a function of time integrating the acceleration:

In Problem 1 you can see in more detail how this integral is solved.

Substituting the initial velocity vector and grouping:

Now we integrate the velocity to obtain the position vector:

The initial position vector r0 is null because the ball starts moving from the origin of coordinates.

Next we are going to write the two components of the position vector:

The truck is moving at a constant speed, therefore the x coordinate of its position as a function of time is given by:

You can see that we have used the same system of coordinates to write the position vector of both the ball and the truck.


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For the ball to fall into the back of the truck, the x-component of the position vectors of both must be equal at the same time:

The time in equation (4) can be calculated from equation (2). When the ball falls into the back of the truck, the y-component of its position vector is zero. Therefore:

And substituting the time in equation (5) into equation (4):

Substituting the givens of the problem statement into this equation and operating you will obtain a second degree polynomial. Its positive root will be v0 (you have to take the positive root because we are calculating the magnitude of a vector, which is always positive).

Finally v0 is:

This is the minimal value of v0 for the ball to fall into the back of the truck (at the rear end).


v0 will have its maximum value when the ball hits the front end of the back of the truck. In this case:

An using the same procedure as before:

Do not forget to include the units in the results.

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