Question

A ball is dropped from the top of a tower of height h. It covers a distance of h/2. In the east second of it's motion. How long does the ball remain in air?

please answer fast..​

Answer 1

Answer:

1.414

Explanation:

In the first second, ball travels h/2 distance with an acceleration of g.

So,    applying      $s =ut+\frac{1}{2}at^{2}$

                            $-h/2=0-1/2 gt^{2}$

                              h = g

Therefore, applying    $s =ut+\frac{1}{2}at^{2}$

                                    $-h =0 -1/2 gt^{2}$

                                      $t= \sqrt{2}$