# Math Help - falling body quadradic function

The general motion equation is given by:

d(t)=1/2 gt^2+v_ot +d_o where:

d(t) is the distance traveled, in metres
g is the acceleration due to gravity in m/s2
t is the time, in seconds (s)
vo is the initial velocity of the moving object
do is the initial height in metres.

An object is dropped from a height of 1,000 m, on Mars, where the acceleration due to gravity is 3.92 m/s2. Find, to two decimal places, the time it will take for the object to hit the surface of Mars.

i get 1.96t^2, is it right?

2. Originally Posted by william
The general motion equation is given by:

d(t)=1/2 gt^2+v_ot +d_o where:

d(t) is the distance traveled, in metres
g is the acceleration due to gravity in m/s2
t is the time, in seconds (s)
vo is the initial velocity of the moving object
do is the initial height in metres.

An object is dropped from a height of 1,000 m, on Mars, where the acceleration due to gravity is 3.92 m/s2. Find, to two decimal places, the time it will take for the object to hit the surface of Mars.

i get 1.96t^2, is it right?
s = 1000m
u = 0
a = 3.92
t = t

We use $s = ut + \frac{1}{2}at^2$

$1000 = 0 + \frac{1}{2}3.92t^2$

$t = \sqrt{\frac{2000}{3.92}}$

edit: yes, the coefficient of $t^2$ is 1.96

3. Hello, William!

Their formula isn't accurate . . .
They should have explained about "upward is positive", etc.

The general motion equation is given by: . $d(t)\:=\:{\color{red}-}\tfrac{1}{2}gt^2+v_ot +d_o$ where:

$d(t)$ is the distance traveled, in metres
$g$ is the acceleration due to gravity in m/s²
$t$ is the time, in seconds,
$v_o$ is the initial velocity of the moving object,
$d_o$ is the initial height in metres.

An object is dropped from a height of 1,000 m, on Mars,
where the acceleration due to gravity is 3.92 m/s².

Find, to two decimal places, the time it will take for the object to hit the surface of Mars.

i get 1.96t^2 .
. . . You have a ${\color{blue}t^2}$ in there?

We are given: . $g = 3.92,\;v_o = 0,\;d_o = 1000$

. . The equation is: . $d(t) \;=\;-1.96t^2 + 1000$

"Hit the surface" means $d = 0\!:\;-1.96t^2 + 1000 \:=\:0\quad\Rightarrow\quad t^2 \:=\:\frac{1000}{1.96}$

. . Therefore: . $t \:=\:\sqrt{\frac{1000}{1.96}} \;\approx\; 22.59\text{ seconds}$