Vertical Motion From A Height

Question

From the roof of a building 25 m tall, a ball is thrown vertically upward with a speed of 20 m/s. Taking g = 10 m/s^2, how long does the ball take to reach the ground below the building?

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This is one-dimensional motion under constant gravitational acceleration, where upward is conveniently chosen as positive. The displacement equation $s = ut + 	frac{1}{2}at^2$ 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 $s = -25	ext{ m}$, with $u = +20	ext{ m/s}$ and $a = -10	ext{ m/s}^2$. 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 $t = 5	ext{ s}$, 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.

Vertical Motion From A Height

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