A ball is thrown upward from the top of a tower 50m high and velocity 20m/s. Find (a) the top height reached from the foot to the tower (b) the total time taken by the object to reached the ground from the momemt it is thrown
Answers
example
You can view this in many ways:
Suppose you are on the rock dropped. Then the rock thrown up will be moving at constant velocity, since gravity pulls it down, but the pseudo force pulls it up. Hence, time to close the distance would be 50/25 = 2sec. Rest is maths.
Suppose you are on the rock thrown up. Then acceleration of dropped stone is again 0. Initial relative velocity is 25 m/s. Time to close the distance is 2 sec.
Suppose you are on center of mass reference frame. Initial cm is at 25 m, with velocity of cm being 12.5 m/s upwards. Forces acting on cm is again gravity, pulling it down. So, pseudo force pulls the balls up. Hence, the balls move at constant velocity wrt cm. Velocity of ball thrown up is 12.5 m/s. Velocity of ball dropped down is 12.5 m/s. Relative velocity is 25 m/s. Distance to travel is 50 m. It takes 2 sec to meet.
Suppose you are on ground reference frame. Then distance travelled by dropped ball is 0.5*g*t^2. Distance travelled by ball thrown up is 25t-0.5*g*t^2. When they meet, s1 + s2 = 50. Solve this quadratic in t, you get 2 sec.
As you can see, choosing a good reference frame simplified the problem considerably