From a pole of height $$10\,m,$$  a stone is thrown vertically upwards with a speed $$5\,m/s.$$  The time taken by the stone, to hit the ground, is $$n$$ times that taken by it to reach the highest point of its path. The value of $$n$$ is [take $$g = 10\,m/{s^2}$$  ]

Solution :
Given, $$H = 10\,m,u = 5\,m/s,g = 10\,m/{s^2}$$
Speed on reaching ground
$$v = \sqrt {{u^2} + 2gH} $$
Now, $$v = u + at \Rightarrow \sqrt {{u^2} + 2gH} = - u + gt$$
Time taken to reach highest point is $$t = \frac{u}{g},$$
$$\eqalign{ & \Rightarrow t = \frac{{u + \sqrt {{u^2} + 2gh} }}{g} = \frac{{nu}}{g}\,\left( {{\text{from question}}} \right) \cr & \Rightarrow 2gH = n\left( {n - 2} \right){u^2} \cr & \Rightarrow n\left( {n - 2} \right) = \frac{{2gH}}{{{u^2}}} = \frac{{2 \times 10 + 10}}{{5 \times 5}} = 8 \Rightarrow n = 4 \cr} $$