# 5.02 Object's dropped from rest

When a stationary object is dropped from above the Earth's surface there is only one direction of motion involved (i.e. the object falls towards the ground). Also the object's initial velocity "u" is zero. Therefore the suvat equations
s = ut + 1/2at2
v = u + at

Simplify to
y =1/2gt2
v = gt

"s" and "a" are replaced with "y" and "g" as before.
However this time there is no minus sign placed before g. This is because we are now taking the downward direction to be positive. This means that "g"" is now positive.
"y" now represents the distance the object has fallen (not its height above the ground as in the previous example) and "v" represents the velocity towards the ground. Choosing the downward direction to be positive simplifies the calculations for these type of problems Example
An object is dropped from 20m
a) How far does the object fall in 1.5 seconds and what is its velocity?
b) How long does it take to reach the ground?

a)
t = 1.5s
g = 9.81 ms-2
y = ?
v = ?

y =1/2gt2
y =1/29.81 x 1.52 = 11.04m
v = gt
v = 9.81 x 1.5 = 14.72 ms-1

b)
t = ?
g = 9.81 ms-2
y = 20

y =1/2gt2     Therefore t = (2y/g)1/2
t = (2 x 20/ 9.81)1/2 = 2.02s