After completing this section you should be able to apply the equations of motion to solve problems involving acceleration due to gravity.
The relationship between forces and acceleration will be covered in the section on dynamics. However in this section we will consider the
acceleration caused by the Earth's gravity. Objects close to the Earth's surface are accelerated at a rate of 9.81 ms -2
(to 3 sig figs).
(By "close to the Earth's surface" we mean distances from the ground up to the heights at which artificial satellites orbit the Earth.)
We give this constant value of acceleration its own symbol g.
It is important to emphasise that in this section we are only considering the acceleration of a free body close to the Earth's surface where
no force other than gravity is acting on it!
We can use the same suvat equations as before to solve problems involving acceleration due to gravity. The most common equations we use are:
s = ut + 1/2at2
v = u + at
However the equations are usually rewritten as
y = ut - 1/2gt2
v = u - gt
The use of "y" instead of "s" is just to emphasise that we are dealing with vertical motion. Also ""g"" replaces "a""
simply to emphasise that it is acceleration due to gravity that we are dealing with. However the change in signs requires further explanation;
Many examples involving acceleration due to gravity will involve a change in direction from upward to downward motion.
It is sensible to choose "+" for up and "-" for down. As g always points downward it will have a negative value i.e.
-9.81. We could write the equation as before (e.g. v = u + gt) and then remember to put -9.81 in for the value of g each time,
or rewrite the equation with the - sign already in it e.g. (v = u -gt) so that we can just insert the magnitude of g (9.81)
each time.
example
An object is launched vertically into the air with an initial velocity of 20 ms -1
what is its velocity and height after:
a) 2 seconds?
b) 4 seconds?
a) t = 2s
v = u -gt v = 20 - 9.81 x 2 = 0.38 ms -1
y = ut - 1/2gt2 y = 20 x 2 - 1/2 x 9.81 x 22 = 20.38m
b) t = 4s
v = u -gt v = 20 - 9.81 x 4 = -19.24 ms -1
y = ut - 1/2gt2 y = 20 x 4 - 1/2 x 9.81 x 42 = 1.52m
The diagram illustrates two sections of the object's motion. First travelling upwards from the ground then falling back again from
its maximum height
At 2 seconds the object is still travelling upwards and its velocity has dropped to 0.38 ms -1 as it approaches its maximum height. At 4 seconds it is returning to the ground with a high (negative) velocity.
