Imagine a motorcycle rider moving with a speed 0.7c past a stationary observer. If the rider tosses a ball in the forward direction with a speed of 0.6c, relative to himself, what is the speed of the ball as seen by the stationary observer? Ans in units of c.
The addition of velocities formula says
(u+v)/(1+uv/c^2)
You can do it all in units of c, so plug in u=0.7 and v=0.6 to get approximately 0.92.
To find the speed of the ball as seen by the stationary observer, we can use the formula for the addition of velocities. The formula is given by (u+v)/(1+uv/c^2), where u is the speed of the motorcycle rider and v is the speed of the ball relative to the motorcycle rider. In this case, the speed of the motorcycle rider is 0.7c and the speed of the ball relative to the rider is 0.6c.
Substituting these values into the formula, we get (0.7c + 0.6c)/(1 + (0.7c)(0.6c) / c^2). Simplifying, we have (1.3c)/(1 + (0.42c^2/c^2)). Continuing to simplify, we have (1.3c)/(1 + 0.42) = (1.3c)/(1.42). Finally, dividing 1.3c by 1.42, the speed of the ball as seen by the stationary observer is approximately 0.92c.