Answer to Question #135409 in Mechanics | Relativity for George

Question

On another planet, a marble is released from rest at the top of a high cliff. It falls 4.00 m in the first 1 s of its motion. Through what additional distance does it fall in the next 1 s?

Answer 1

Distance traveled in first second $=s=ut+ \frac{1}{2}at^2=\frac{1}{2}at^2$ $($ As $u=0)$

As $s_1= 4 \ m$ for $t=1\ s$

So, $\frac{1}{2}a\times 1^2= 4\implies a= 8 \ m/s^2$

Now ,to find the distance traveled in first two seconds.

$t= 2 \ s, a= 8\ m/s^2$

$s_2= \frac{1}{2}at^2= \frac{1}{2}\times 8\times 2^2= 16\ s$

So distance traveled in $2$nd second is the difference between distance traveled by particle in first two seconds and first second.

Required distance$= s_2-s_1= 16-4=12\ m$