A stone is projected vertically up from the ground with velocity 40 ms^–1. The interval of time between the two instants at which the stone is at a height of 60 m above the ground is (g = 10 ms^–2)

At h = 60 m velocity v² - u² =  2as

v² = 40² - 2 x 10 x 60 = 1600 - 1200

v² = 400 $\Rightarrow$ v = 20 m/s

For the required time interval, we can calculate by assuming it as time of flight for a body projected vertically upwards with initial velocity of 20m/s

$\therefore \mathrm{t}=\frac{2\mathrm{v}}{\mathrm{g}}=\frac{2\times 20}{10}=4\mathrm{s}$