Free Fall — Time to Reach Ground from Height h

Free fall is the cleanest application of the second equation of motion — initial velocity is zero, acceleration is $g$, and displacement equals height. The only way to get this wrong is a sign error or forgetting that a dropped stone starts from rest. Let's make sure neither happens.

Question

A stone is dropped from the top of a building 80 m high. Find:

1. The time taken to reach the ground
2. The velocity with which it strikes the ground

(Take $g = 10$ m/s²)

Solution — Step by Step

Step 1: Write the given information clearly.

• Initial velocity: $u = 0$ m/s (dropped, not thrown)
• Height (displacement): $h = 80$ m (downward)
• Acceleration: $g = 10$ m/s² (downward)
• Find: time $t$ and final velocity $v$

Sign convention: Taking downward as positive throughout.

Step 2: Find the time using the second equation of motion.

$s = ut + \frac{1}{2}gt^2$

Since $u = 0$:

$80 = 0 + \frac{1}{2} \times 10 \times t^2$

$80 = 5t^2$

$t^2 = 16$

$t = 4 \text{ s}$

Step 3: Find the final velocity.

Using the first equation of motion:

$v = u + gt = 0 + 10 \times 4 = 40 \text{ m/s}$

Or using the third equation (without needing time):

$v^2 = u^2 + 2gh = 0 + 2 \times 10 \times 80 = 1600$

$v = 40 \text{ m/s}$

Answers:

• Time to reach ground $=$ 4 seconds
• Velocity at ground $=$ 40 m/s (downward)

Why This Works

When a body is dropped (released from rest), it starts with zero velocity and gains speed every second due to gravity. In each second, it gains 10 m/s (when $g = 10$ m/s²). After 4 seconds, velocity $= 40$ m/s. Clean.

The distance doesn't increase uniformly — it increases as $t^2$. In the first second the stone falls 5 m, in the second second it falls 15 m more (total 20 m), in the third it falls 25 m more (total 45 m), in the fourth it falls 35 m more (total 80 m). This non-uniform falling is why Galileo's discovery was revolutionary — heavier objects don't fall faster, all objects fall the same way.

Alternative Method — Using the Direct Formula

For a body dropped from rest, we derived:

$h = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2 \times 80}{10}} = \sqrt{16} = 4 \text{ s}$

And: $v = gt = 10 \times 4 = 40$ m/s

Or: $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 80} = \sqrt{1600} = 40$ m/s

Both routes confirm the same answer. For MCQs, the direct formula is fastest.

Extension — What if the Stone is Thrown Downward?

If the stone is thrown downward at 10 m/s (instead of just dropped), then $u = 10$ m/s downward.

$80 = 10t + \frac{1}{2} \times 10 \times t^2 = 10t + 5t^2$

$5t^2 + 10t - 80 = 0$

$t^2 + 2t - 16 = 0$

Using quadratic formula: $t = \dfrac{-2 + \sqrt{4 + 64}}{2} = \dfrac{-2 + \sqrt{68}}{2} \approx \dfrac{-2 + 8.25}{2} \approx 3.1$ s

Makes sense — throwing it downward with initial velocity means it reaches the ground faster than 4 seconds.

Common Mistake