Answer to Question #136550 in Mechanics | Relativity for Dominik

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

Solution

Displacement in 1sec can be given by second equation of motion

So

$S_1=ut+\frac{1}{2}at^2$

And

S₁=4m t=1sec u=0m/sec

So

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

So now displacement in 2sec

$S_2=\frac{1}{2}at_2^2$

$S_2=\frac{1}{2}\times8\times2^2\\=16m$

Displacement t=1 and t=2 sec time interval

$S_3=S_2-S_1$

$S_3=16-4=12m$

So additional distance traveled in the next 1sec is 12m .