A particle is projected from ground with velocity $20\sqrt{2} \mathrm{m}/\mathrm{s}$ at $45°$. At what time, the particle is at height 15 m from ground? $(Take, g=10 \mathrm{m}/{\mathrm{s}}^2)$

Vertical component of initial velocity,  $u_y=20\sqrt{2}\sin 45°=20 \mathrm{m}/\mathrm{s}$

Vertical component of velocity at height h from the ground,  $v_y=\pm \sqrt{{u_y}^2-2gh}$

Time taken for the vertical component to reduce from $u_y$ to $v_y$,  

$t=\frac{u_y-v_y}{g}=\frac{u_y\pm \sqrt{{u_y}^2-2gh}}{g}$.

When h = 15 m,  $t=\frac{20\pm \sqrt{{20}^2-\left(2\times 10\times 15\right)}}{10}=2\pm 1 \text{s}$

$\therefore t=3 \mathrm{s} \text{and} 1 \mathrm{s}.$