How far does a free-falling object fall in the $3$rd second?

Question

The question:

An object is dropped from a cliff. How far does the object fall in the 3rd second?"

I calculated that a ball dropped from rest from a cliff will fall $45\text{ m}$ in $3 \text{ s}$, assuming $g$ is $10\text{ m/s}^2$.

$$s = (0 \times 3) + \frac{1}{2}\cdot 10\cdot (3\times 3) = 45\text{ m}$$

But my teacher is telling me $25\text{ m}$!

EDITS: His reasoning was that from $t=0$ to $t=1$, $s=10\text{ m}$, and from $t=1$ to $t =2$, $s=20$...

The mark scheme also says $25\text{ m}$

Answer 1

Your teacher is correct. Question asks how much distance is covered between $t= 2$ and $ t= 3.$ Time lapse is 1 second, that is, in the third second of duration. In meters, distance travelled =

$$ s = \frac12 \cdot 10\cdot (3^2-2^2) = 25, $$

and, if you draw the parabola graph, $s_2-s_1 = a (t_2^2-t_1^2)/2. $

Answer 2

We have

$$h=\frac {1}{2}gt^2+v_0t+h_0$$

$$=\frac {1}{2}gt^2$$

$$=\frac {1}{2}.10. (3)^2=45 m $$.

Answer 3

You are correct. If the question is exactly how you phrased it, the displacement should be 45m down.

We know that:

a (acceleration) = $10m/s^2$

t (time) = 3 seconds

u (initial velocity) = $0m/s$

Hence, using the formula $s=ut-(1/2)at^2$ the answer should be 45 m.

Answer 4

Using the formula $s=ut+1/2at^2$

Where $a$ is acceleration, $u$ is the initial velocity, $t$ is the time and $s$ the the displacement.

We can deduce that the displacement will be $45m$.

You are indeed correct!