1. The problem statement, all variables and given/known data Imagine that you are flying on an airliner on a long flight to Europe at a constant speed of 300 m/s a) you throw a ball towards the back of the plane at 20 m/s. You then shine a beam of light towards the back of the plane. How will these two things i) the ball and ii) the light-- appear to move from the Earth's frame of reference b) would you expect your watch to be affected by time dilation? 2. Relevant equations Δtm = Δts/√(1-v2/c2) V(be)= -V(bp) + V(pe) 3. The attempt at a solution a) i) b= ball; e= earth; p= plane ∴ V(be)= -V(bp) + V(pe)= -20 m/s + 300 m/s = 280 m/s [forward]. ii) The light will remain at a constant 3.0 x 10^8 m/s from the earth’s frame of reference. Velocity is dependent on its’ inertial reference frame and its’ direction of travel. From the earth frame of reference, both the direction of the thrown ball and the direction of the airplane travel determines its’ velocity as per vector subtraction. If we were to look at the ball moving from a different inertial frame of reference (i.e. from the plane), the ball would experience only the velocity of 20m/s [backwards] and if the ball was thrown forward, the velocity of the ball would be 20 m/s + 300 m/s= 320 m/s [forward]. Light on the other hand is independent of its’ reference frame and shows a constancy of 3.0 x 10^8 m/s relative to all inertial reference frames (special relativity postulate 2), regardless of which direction it is travelling. If we were to measure the speed of light from inside the plane or if it were travelling back or forwards, its’ velocity would still be 3.0 x 10^ 8 m/s. b) If time was being measured from the earth’s frame of reference, time dilation would still occur relative to the time experienced from the airplane. However, because the plane is not travelling at a speed close enough to the speed of light, increasing the time difference between both reference frames, time difference and time dilation would be negligible. As such, my watch would not be accurate enough to pick up the time difference. If time was measured from the inside the plane, my watch would only experience normal time and would not experience time dilation at all. Could someone review my answers and tell me if I am understanding these concepts? Thank you for your time!