CK-12-Physics-Concepts - Intermediate

2.7. Acceleration Due to Gravity http://www.ck12.org

Solution: We are looking for displacement and we have time and acceleration. Therefore, we can used=^12 at^2.

Displacement after 1.00 s=

( 1

2

)

( 9 .80 m/s^2 )( 1 .00 s)^2 = 4 .90 m

Displacement after 2.00 s=

( 1

2

)

( 9 .80 m/s^2 )( 2 .00 s)^2 = 19 .6 m

Displacement after 3.00 s=

( 1

2

)

( 9 .80 m/s^2 )( 3 .00 s)^2 = 44 .1 m

Example: (a) A person throws a ball upward into the air with an initial velocity of 15.0 m/s. How high will it go
before it comes to rest? (b) How long will the ball be in the air before it returns to the person’s hand?

Solution: In part (a), we know the initial velocity (15.0 m/s), the final velocity (0 m/s), and the acceleration (-9.80
m/s^2 ). We wish to solve for the displacement, so we can usevf^2 =vi^2 + 2 adand solve ford.

d=vf

(^2) −vi 2
2 a =
( 0 m/s)^2 −( 15. 0 m/s)^2
( 2 )( 9. 80 m/s^2 )
= 11 .5 m
There are a number of methods by which we can solve part (b). Probably the easiest is to divide the distance traveled
by the average velocity to get the time going up and then double this number since the motion is symmetrical –that
is, time going up equals the time going down.
The average velocity is half of 15.0 m/s, or 7.5 m/s, and dividing this into the distance of 11.5 m yields 1.53 sec.
This is the time required for the ball to go up and the time for the ball to come down will also be 1.53 s, so the total
time for the trip up and down is 3.06 sec.
Example: A car accelerates with uniform acceleration from 11.1 m/s to 22.2 m/s in 5.0 s. (a) What was the
acceleration and (b) how far did it travel during the acceleration?
Solution:
(a) a=∆∆vt=^22.^2 m/s 5 .− 011 s.^1 m/s= 2 .22 m/s^2
(b) We can find the distance traveled byd=vit+^12 at^2 , or we could find the distance traveled by determining the
average velocity and multiply it by the time.
d=vit+

1

2

at^2

= ( 11 .1 m/s)( 5 .0 s)+

(

1

2

)

( 2 .22 m/s^2 )( 5 .0 s)^2

= 55 .5 m+ 27 .8 m

=83 m

d= (vave)(t) = ( 16 .6 m/s)( 5 .0 s) =83 m

Example: A stone is dropped from the top of a cliff. It is hits the ground after 5.5 s. How high is the cliff?

Solution:

d=vit+^12 at^2 = (0 m/s)( 5 .5 s)+

( 1

2

)

( 9 .80 m/s^2 )( 5 .5 s)^2 =150 m

Summary

• At any given location on the earth and in the absence of air resistance, all objects fall with the same uniform
  acceleration.


• We call this acceleration the acceleration due to gravity on the earth and we give it the symbolg.


• The value ofgis 9.80 m/s^2.