1. Objects Thrown Vertically Upward – Free Fall 2

2. Sign Convention: A Ball Thrown Vertically Upward
Displacement is positive (+) or negative (-) based on LOCATION.
v = -
UP = +
v = -
a = -
y = 0
y = 0
Release Point
Velocity is positive (+) or negative (-) based on direction of motion.
y = -Negative
v= -Negative
a = -
Acceleration due to gravity is ALWAYS (-) because the direction of force (weight) is down.

3. + Step 1. Draw and label a sketch.
Example 1: A ball is thrown vertically upward with an initial velocity of 30 m/s. What are its position and velocity after 2 s, 4 s, and 7 s?
Step 1. Draw and label a sketch. +
a = g
Step 2. Indicate + direction and force direction.
Step 3. Write down given and the quantities to find.
vi = +30 m/s
a = m/s t = 2, 4, 7 s
vi = + 30 m/s y = ? vf = ?

4. Finding Displacement:
a = g +
vi = +30 m/s
Step 4. Select equation that contains y (d) and not vf .
y = (30 m/s)t + ½(-9.81 m/s2)t2
Substitution of t = 2, 4, and 7 s will give the following values:
y = m; y = m; y = m

5. Finding Velocity:
a = g +
vi = +30 m/s
Step 5. Find v from equation that contains v and not y:
vf = 30 m/s+(-9.81 m/s2)t
Substitute t = 2, 4, and 7 s:
vf = m/s; vf = m/s; vf = m/s

6. Example 1: (Cont.) Now find the maximum height attained:
a = g +
vi = +30 m/s
Displacement is a maximum when the velocity vf is zero.
To find ymax we substitute
t = 3.06 s into the general equation for displacement:

7. Example 1: (Cont.) Finding the maximum height:
ymax = (30 m/s)t + ½(-9.81 m/s2)t2
for t = 3.06 s
a = g +
vi =+30 m/s
Substituting with units, we obtain:
y max= 91.8 m m
ymax = 45.9 m
How would you find the maximum height differently?

8. Example 1: (Cont.) Finding the maximum height:
a = g +
vi =+30 m/s
Substituting with units for t = 3.06 s, we obtain:
ymax = 45.9 m