Kinematics: Common Mistakes and Fixes (3)

Question

A particle is projected vertically upwards with speed $u = 20$ m/s from the top of a tower of height $25$ m. Find the time taken to hit the ground. Take $g = 10$ m/s$^2$.

This looks like a one-line substitution, but every year students lose marks here because of a sign convention slip. Let’s walk through it carefully and then talk about the three places where students mess up.

Solution — Step by Step

Why This Works

The trick is that we are not asking “when does it come back to the start?” — we are asking when it reaches a point $25$ m below the start. So $s$ has to carry a negative sign. Once that’s locked in, the algebra is just a standard quadratic.

The negative root $t = -1$ s has a meaning, by the way: it’s the time at which a particle would have left the ground to reach the projection point with the given velocity. We discard it because the motion only started at $t = 0$.

Alternative Method

Split the motion into two phases.

Phase 1 (going up to the highest point above the tower): $v = 0$, so time $t_1 = u/g = 2$ s and height gained $h = u^2/(2g) = 20$ m.

Phase 2 (free fall from height $25 + 20 = 45$ m): $45 = \tfrac{1}{2}(10)t_2^2 \implies t_2 = 3$ s.

Total time $= t_1 + t_2 = 2 + 3 = 5$ s. Same answer.

Common Mistake

Students write $25 = 20t - 5t^2$ (positive $25$) and end up with $5t^2 - 20t + 25 = 0$, whose discriminant is $400 - 500 < 0$ — no real roots. They panic, redo the question, and lose four minutes. The root cause is always the same: not committing to a sign convention upfront.