Galilean Relativity: I can't figure this out

[motherlovebone]

Here is the problem: "A passenger in a convertible throws a ball up into the air. The car is going 35.0 m/s. The upward velocity of the ball is 8.00 m/s. Give the equations which specify the ball position at any given time with respect to the passenger (prime values) and respect to an observer on the road (un prime values)."

Here is my problem . . . The teacher told us that in (x, y, z) and (x', y', z'), y and y' are the same. He told us that x = x' + vt and x' = x - vt. However, we weren't given t or t' so I cannot find x or x', and I absolutely have no clue how to find z or z'. Can someone give me hints or something? Thanks!

 

[dicerandom]

You need to work out for yourself what path the ball will take given the initial conditions. I reccomend simply declaring that the ball was thrown up at t=0, at which point x=x'=y=y'=z=z'=0.

 

[kyle8921]

I understand your confusion and I am happy to provide some guidance on this problem.

Firstly, it is important to understand the concept of Galilean relativity. This principle states that the laws of physics are the same for all observers in uniform motion. In other words, the laws of physics do not change depending on the frame of reference or the observer's perspective.

Now, let's apply this principle to the problem at hand. We have two reference frames - one for the passenger in the car (x', y', z') and one for the observer on the road (x, y, z). As mentioned by your teacher, the y and y' coordinates will be the same for both frames because the ball is moving only in the horizontal direction.

To find the equations for the ball's position, we can use the equation x = x0 + vt, where x0 is the initial position, v is the velocity, and t is the time. In this case, we can use x' and x to represent the ball's position in the passenger's frame and the observer's frame, respectively.

For the passenger's frame, we know that the initial position of the ball is 0 (since it is thrown from the passenger's hand) and the velocity is 8.00 m/s. Therefore, the equation for the ball's position in the passenger's frame would be x' = 8.00t.

For the observer's frame, we need to take into account the motion of the car. The car is moving at a constant velocity of 35.0 m/s in the positive x direction. This means that the initial position of the ball in the observer's frame is -35.0t (since the car has already moved 35.0 meters while the ball is in the air) and the velocity is 8.00 m/s. Therefore, the equation for the ball's position in the observer's frame would be x = -35.0t + 8.00t = -27.0t.

To find the equations for z and z', we can use the same approach. We know that the initial position in the z direction is 0 for both frames, and the velocity in the z direction is 0 since the ball is only moving in the horizontal direction. Therefore, the equations for z and z' would both be 0.

I hope this explanation helps you understand the problem better. Remember