Acceleration due to gravity is so important that its magnitude is given its own symbol,
g . It is constant at any given location on Earth and has the average value
Although
g varies from 9.78 m/s
2 to 9.83 m/s
2 , depending on latitude, altitude, underlying geological formations, and local topography, let’s use an average value of 9.8 m/s
2 rounded to two significant figures in this text unless specified otherwise. Neglecting these effects on the value of
g as a result of position on Earth’s surface, as well as effects resulting from Earth’s rotation, we take the direction of acceleration due to gravity to be downward (toward the center of Earth). In fact, its direction
defines what we call vertical. Note that whether acceleration
a in the kinematic equations has the value +
g or −
g depends on how we define our coordinate system. If we define the upward direction as positive, then
and if we define the downward direction as positive, then
.
One-dimensional motion involving gravity
The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So, we start by considering straight up-and-down motion with no air resistance or friction. These assumptions mean the velocity (if there is any) is vertical. If an object is dropped, we know the initial velocity is zero when in free fall. When the object has left contact with whatever held or threw it, the object is in free fall. When the object is thrown, it has the same initial speed in free fall as it did before it was released. When the object comes in contact with the ground or any other object, it is no longer in free fall and its acceleration of
g is no longer valid. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude
g . We represent vertical displacement with the symbol
y .
Kinematic equations for objects in free fall
We assume here that acceleration equals −
g (with the positive direction upward).
Problem-solving strategy: free fall
Decide on the sign of the acceleration of gravity. In
[link] through
[link] , acceleration
g is negative, which says the positive direction is upward and the negative direction is downward. In some problems, it may be useful to have acceleration
g as positive, indicating the positive direction is downward.
Draw a sketch of the problem. This helps visualize the physics involved.
Record the knowns and unknowns from the problem description. This helps devise a strategy for selecting the appropriate equations to solve the problem.
Decide which of
[link] through
[link] are to be used to solve for the unknowns.
Free fall of a ball
[link] shows the positions of a ball, at 1-s intervals, with an initial velocity of 4.9 m/s downward, that is thrown from the top of a 98-m-high building. (a) How much time elapses before the ball reaches the ground? (b) What is the velocity when it arrives at the ground?
The positions and velocities at 1-s intervals of a ball thrown downward from a tall building at 4.9 m/s.
Strategy
Choose the origin at the top of the building with the positive direction upward and the negative direction downward. To find the time when the position is −98 m, we use
[link] , with
.
Solution
Substitute the given values into the equation:
This simplifies to
This is a quadratic equation with roots
. The positive root is the one we are interested in, since time
is the time when the ball is released at the top of the building. (The time
represents the fact that a ball thrown upward from the ground would have been in the air for 5.0 s when it passed by the top of the building moving downward at 4.9 m/s.)
For situations when two roots are obtained from a quadratic equation in the time variable, we must look at the physical significance of both roots to determine which is correct. Since
corresponds to the time when the ball was released, the negative root would correspond to a time before the ball was released, which is not physically meaningful. When the ball hits the ground, its velocity is not immediately zero, but as soon as the ball interacts with the ground, its acceleration is not
g and it accelerates with a different value over a short time to zero velocity. This problem shows how important it is to establish the correct coordinate system and to keep the signs of
g in the kinematic equations consistent.