Vertical Motion From A Height

This is one-dimensional motion under constant gravitational acceleration, where upward is conveniently chosen as positive. The displacement equation $s=ut+\frac{1}{2}a{t}^2$ applies throughout the flight, including the upward and downward portions, provided signs are kept consistent. The ball lands 25 m below its launch point, so its final displacement is $s=-25\text{ m}$, with $u=+20\text{ m/s}$ and $a=-10{\text{ m/s}}^2$. Substituting gives $-25=20t-5t^2$, which rearranges to $5t^2-20t-25=0$, or $t^2-4t-5=0$. Factoring yields $(t-5)(t+1)=0$, so $t=5\text{ s}$, discarding the negative root. The option 4 s is wrong because it ignores the building height and only returns the ball to roof level after $2u/g=4$ s. The option 2 s gives only the time to reach the peak. The option 6 s overestimates by mis-signing the displacement. This uses the NCERT single-equation approach to projectile-up-then-down motion. As a check, $t=5$ s exceeds the 4 s round trip to roof level, which is required since the ball must then fall the extra 25 m.