A rugby player (A) moves with a constant velocity vA with respect to a reference frame at rest with origin in O. He throws the ball towards another player (B) who runs behind him with constant velocity vB = 9 i (m/s) with respect to the same reference frame (see figure) so that player B catches the ball. The initial velocity of the ball is v’0 = -12 i + 5 j (m/s) with respect to A. The said initial velocity is v0 = -5 i + 5 j (m/s) with respect to O.
- Determine player’s A velocity vA with respect to O.
- What is the acceleration of the ball with respect to player A? And with respect to O?
If player A throws the ball from an initial height y0 = 1.6 m, determine:
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- the initial distance x0 between the two players so that player B catches the ball at a height y = 1.3 m.
- the velocity vector of the ball with respect to O at the moment when player B catches the ball.
- the tangential acceleration and normal acceleration of the ball at the highest point of its trajectory.
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Solution:
In this problem the two reference systems, the one that is resting on the ground (O) and the player A are inertial. We will therefore use Galilean transformations to determine the velocity of player A with respect to O. The general expression of Galilean transformations for velocities is:
Where V is the velocity of the reference frame in constant relative motion (in this case it is the velocity vA of player A) with respect to O.
We can then apply the above equation using the initial velocity of the ball calculated by both observers:
And after replacing both velocities we get:
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The acceleration of the ball for both observers is that of gravity g; since both are inertial, any object has the same acceleration for them.
For player B to catch the ball the x coordinate of the position vector of both must be the same measured with respect to O. We will therefore write the equation of the x coordinate of the position vector of player B and that of the ball so that we can later equate them.
Ball:
The ball describes a parabolic motion. Therefore, the x and y coordinates of its position vector using the axes represented in the figure are:
And after substituting the data provided in the problem statement we get:
On the other hand, the distance traveled by player B is given by:
By equating both we get:
To determine how long the ball is in the air we use the givens of the statement which tells us that player B catches the ball at a height y = 1.3 m from the ground:
And finally after substituting in the expression of x0 we get:
To determine the velocity of the ball with respect to O at the instant when player B catches it, we first write the components of its velocity vector as a function of time:
As we have previously calculated how long the ball is in the air (t = 1.05 s), we can substitute it in the components of the ball’s velocity:
The trajectory of the ball as well as the axes tangent (t) and perpendicular (n) to the trajectory at its highest point are represented in the following figure. The total acceleration of the ball is that of gravity (since it has a parabolic motion).
The total acceleration has no projection on the axis tangent to the trajectory at its highest point, so the tangential acceleration is zero at this point. Since the total acceleration (g) is on the axis perpendicular to the trajectory, the normal acceleration will be equal to the acceleration of gravity.
In this problem we used g = 10 m/s2 .
Do not forget to add the units to the results of the problem.
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