Question

A stone thrown vertically upwards with
initial velocity u reaches a height 'h'
before coming down. Show that the
time taken to go up is same as the time
taken to come down.
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Answer 1

Answer:

Explanation:  Distance covered and taken time when going up

                      v = u + at

                       t = u/g            (v = 0)             -------------       (1)

                       s = ut + 1/2 at²

                      h = u×u/g - 1/2 g(u/g)²

                      h = u²/2g

                      time taken when going down

                      s = ut + 1/2 at²

                      u²/2g = 1/2 gt²           ( u = 0)

                      t² = (u/g)²

                      t = u/g                                  -------------       (2)

                    From equation (1) and equation (2) its clear then time taken is                                        

                     same in both scenarios.