A body thrown vertically up to reach its maximum height in t second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is

Step-by-step solution

The ball is thrown vertically upwards then according to equation of motion (0)^2-u^2=-2 g h and 0=u-g t From Eqs. (i) and (ii), h= g t^2 2 When the ball is falling downwards after reaching the maximum height aligned s & =u t^ + 1 2 g (t^ )^2 h 2 & =(0) t^ + 1 2 g (t^ )^2 t^ & = h g t^ & = t 2 aligned Hence, the total time from the time of projection to reach a point at half of its maximum height while returning =t+t^ =t+ t 2