Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

1. Imagine two inertial frames, S and S\'. Inertial frame S\' moves with velocit

ID: 1652410 • Letter: 1

Question

1. Imagine two inertial frames, S and S'. Inertial frame S' moves with velocity v_0= 5 m/s in the upward (positive y) direction as seen by an observer in frame S. Now imagine that a person at rest in frame S throws a ball with mass m straight up into the air with initial velocity v_0 (with respect to frame S) from an initial height of y=y_0=0 at time t=0. Use Galilean relativity, and take gravity into account. Use a reasonable number of signi cant digits and appropriate units. (Questions follow.)

a) According to frame S', does the ball appear to be moving after it is thrown for t >0?

b) What is the acceleration of the ball d^2y/dt^2 in frame S? And what is the acceleration of the ball d^2y'/dt^2 in frame S'? Give numbers with appropriate units.

c) What is the maximum height y_max that the ball reaches according to frame S?

d) What is the position of the ball y'_ymax according to frame S' when it reaches its maximum height y_max in frame S?

Explanation / Answer

1. given two frames of references, S and S' ( both inertial)
   S' moves with Vo = 5 m/s in +y direction as seen by observer in S

   Now a person in S throws a ball in air , initial velocity Vo, initial height, yo = 0, at t = 0

   a. For a person in S', the ball is always moving along -ve y axis, never having a +ve velocity, infact he observes the ball to have ever increasing -ve velocity after it started from rest
   b. in frame S, acceleration of ball is -g [ in -ve y direction]
   As S' is also inertial frame, acceleration of ball is -g 9 -n -ve y direcrtion)

   c. in frame S
       hmax = ?
       2*hmax*g = Vo^2
       hmax = V^2/2g = 25/2g = 1.274 m
   d. time taken to reach hmax = t
       Vo = gt
       t = Vo/g = 5/g = 0.509 s
       distance covered by S' accoring to S in this time = Vo*t = 2.54 m

       so, h_max according to S' = 1.274-2.54 = -1.2744 m