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 $25\text{ m}$
$\endgroup$ 44 Answers
$\begingroup$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. $
$\endgroup$ 4 $\begingroup$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.
$\endgroup$ 2 $\begingroup$We have
$$h=\frac {1}{2}gt^2+v_0t+h_0$$
$$=\frac {1}{2}gt^2$$
$$=\frac {1}{2}.10. (3)^2=45 m $$.
$\endgroup$ 6 $\begingroup$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!
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