**Problem Statement:**

A boy at rest (*O*) throws a ball from a pier of height h = 10 m. The initial velocity of the ball has a magnitude v_{0} = 12 m/s and a direction α = 60^{0} (see figure). A second frame of reference (*O’*) is on a boat that has a constant velocity whose magnitude is v_{b} = 6 m/s with respect to *O*.

- Calculate the position and velocity vectors of the ball with respect to
*O*. - Calculate the velocity vector of the ball with respect to
*O’*. - What trajectory does the drop describe with respect to
*O*? And with respect to*O’*? - Calculate the acceleration of the ball with respect to both frames of reference
- Calculate the horizontal distance from the boy at which the ball hits the water.
- Calculate the horizontal distance that the ball has traveled with respect to the boat at this moment.
- If the boat has an acceleration
**A**= 4**i**, what is the acceleration of the ball with respect to it?

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**Solution:**

Since the ball is initially at the origin of coordinates, its initial position vector is null. Its acceleration is that of gravity. Taking into account the signs of the axes, its acceleration vector is given by:

Where we have used g = 10 m/s^{2}.

The components of the initial velocity vector can be calculated from its magnitude and direction. In this problem you can see how.

The velocity vector of the ball as a function of time is calculated integrating the acceleration vector. If you want to see in more detail how this integral is done, check out this problem.

Integrating the expression above and substituting the initial velocity vector:

The position vector can be calculated integrating the velocity vector:

Integrating:

The boat (*O’*) is an inertial reference frame with constant relative motion with respect to *O*. Therefore we are going to use the Galilean transformations to find the velocity vector of the ball with respect to it:

**V** is the velocity of the reference frame *O’* with respect to *O*; in this problem it is the velocity of the boat **v _{b}** = 6

**i**.

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With respect to *O* the ball follows a parabolic trajectory and with respect to *O’* a ver tical trajectory because, as you can see in the equation above, its velocity only has vertical component with respect to this reference frame.

The ball has the same acceleration for both reference frames because they are inertial. This acceleration is that due to gravity.

When the ball hits the water its distance from the pier can be determined using its position vector with respect to *O*. We can find below its components:

At this moment the *y* component of its position vector will be -h. Substituting in the equation of *y* we obtain a second degree polynomial. Taking its positive root we obtain the flight time of the ball.

La distancia al embarcadero se obtiene sustituyendo este tiempo en la ecuación de la coordenada *x* del vector de posición de la pelota:

Since the ball follows a vertical trajectory with respect to the boat, the horizontal distance from it will be zero.

If the boat has an acceleration **A**, *O’* will be a **non-inertial reference frame**. Therefore the acceleration of the ball for him will be:

And the trajectory of the ball will no longer be a straight line.

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