Gravity, velocity time and distance problem

[-Physician]

Homework Statement

A rock from high of $h=180m$ goes down and the gravity is $g=9.81\frac{m}{s^2}$.
Find for how much time did the rock fall down and the velocity of the rock.

Homework Equations

$v^2=2gh$
$v=\sqrt{2gh}$
$t=\frac{v}{g}$

The Attempt at a Solution

First of all we will find the velocity because, to find time we need the velocity so:
$v=\sqrt{2gh}=\sqrt{2χ9.81\frac{m}{s^2}χ180m}= \sqrt{3531.6}=59.43 \frac{m}{s}$
$t=\frac{v}{g}=\frac{59.43\frac{m}{s}}{9.81\frac{m}{s^2}}=6.1s$
is that correct way to do it?
BTW. how to make the square root on formulas?

 

[LawrenceC]

What you have done is correct. Be careful with your significant figures. You provide 4 significant figures for the velocity yet only use 3 for g.

I don't use formulas so I cannot help you there.

 

[-Physician]

Thanks, but i have two more questions:
Where do we know that $t=\frac{v}{g}$, if in formula of velocity $v=√2gh$ we don't have time $t$?
and would gravity formula be $g=\frac{v^2}{h}$ or $g=\frac{v}{t}$ and $v=gt$?

 

[LawrenceC]

The parameter g is the acceleration of gravity in a free fall. In other words if you drop a ball, it accelerates toward the ground at a rate of 9.81 m/sec^2. The velocity attained during a free fall is (with no air drag)

v = gt

Rearranging,

t=v/g


As for v=sqrt(2gh), that formula can get derived a least a couple of ways.

For an object dropped with no initial velocity, the distance it falls is

h = 0.5 g t^2

But we know that

v = g t

so t = v/g

Substitute for t in the displacement equation

h = 0.5 g (v/g)^2 = 0.5 v^2/g

Therefore

v = sqrt(2gh)

The same formula can be derived from conservation of energy considerations.

 

[-Physician]

got it now thanks!