STEPS TO SOLVE PROBLEMS WITH A V0Y
1. Draw a picture
2. Split the problem in 2 parts, (where the object stops
and starts to fall back down)
3. Label all that we know and what we want to know
4. Figure out which K equation to use (Use Parabolic
graph to help simplify situation)
5. Plug and Chug
6. Write answer and circle it. Check to see if the answer
is reasonable
PARABOLIC MOTION
The height is the same at the
same time from Vy=0.
The Velocities are equal with
opposite directions at these
Points as well
Vy=0
EXAMPLE # 1
Mobley throws a rock straight up in the air with an initial velocity of 8m/s.
a) How high does it go? b) how long is it in the air?
Vy=0
V0y=8 m/s
Step 3 Label
Know
V0y= 8 m/s
Vfy= 0 m/s
ay= -9.81 m/s
Need
∆y=?
t=?
Know
V0y= 10 m/s
Vfy= 0 m/s
ay= -9.81 m/s
Need
∆y=?
t=?
Pick the right equation:
For part A) v2 = v02 + 2a(Δy)
For Part B) v = v0 + aΔt
For part A) v2 = v02 + 2a(Δy)  0= (102)+2(-9.81) Δy  (19.62) Δy = 100 
100/19.62 = Δy  Δy = 5.10m
For Part B) v = v0 + aΔt  0 = 10 +(-9.81) Δt  9.81Δt = 10  Δt = 10/9.81 
Δt = 1.01s
EXAMPLE # 2
You want to throw a ball straight up into the air so that it
reaches a height of 3.3 m above the ground.
a) If the ball leaves your hand at a height of 1.2 m above
the ground, how fast do you need to throw it?
b) How long does it take the ball to hit the ground? You
may ignore any effects of air resistance throughout this
problem.
Step # 1
Draw and
Label
You want to throw a ball straight up into the air so that it reaches a height of 3.3 m
above the ground.
a) If the ball leaves your hand at a height of 1.2 m above the ground, how fast
do you need to throw it?
b) How long does it take the ball to hit the ground? You may ignore any effects
of air resistance throughout this problem.
Know
Part 1
Vfy = 0 m/s
∆y=3.3m - 1.2m
ay = -9.81 m/s2
Part 2
V0y = 0 m/s
∆y=3.3m
ay = -9.81 m/s2
Need to Know
Part 1
V0y = ?
t=?
Part 2
t=?
Know
Need to Know
Part 1
Part 1
Vfy = 0 m/s
V0y = ?
∆y=3.3m - 1.2m
ay = -9.81 m/s2
Part 2
Part 2
V0y = 0 m/s
t1+ t2 = Total Time = ?
∆y=3.3m
ay = -9.81 m/s2
Choose which K Equation to use
1. y = y0 + v0Δt + ½ a(Δt)2 (relates position and time)
2. v = v0 + aΔt
(relates velocity and time)
3. v2 = v02 + 2a(Δy)
(relates velocity and position)
For Part A) I would choose equation 3 to start with.
For part B) I would Choose Equation 1
PLUG AND CHUG PART A
PLUG AND CHUG PART B OPTION 1
Solve for t1 and then t2
Know
Need to Know
Part 1
Part 1
Vfy = 0 m/s
t1 = ?
∆y=3.3m - 1.2m
ay = -9.81
So t1 + t2 = .82 +.65
 1.47 ~ 1.5
m/s2
V0 = 6.4 m/s
Part 2
V0y = 0 m/s
Part 2
t2 =?
∆y=3.3m
ay = -9.81 m/s2
For t1 use v = v0 + aΔt 0=6.4-9.81(t1) 9.81(t1) =6.4 
t1 = 6.4/9.81  t1 = .65
For t2 use 1. ∆y = v0Δt + ½ a(Δt)2  3.3 = 0 +1/2(-9.81) (Δ t2)2 
(-3.3)2=(-9.81) (Δ t2)2 -6.6/-9.81= (Δ t2)2 
6.6
9.81
= Δ t2  Δ t2 = .82
PLUG AND CHUG PART B OPTION 2
CHECK TO SEE IF REASONABLE AND WE
UNDERSTOOD PART A
v22 = v12 + 2a(y2 – y1)
0 = v12 + 2(-9.8 m/s2)(3.3 m – 1.2 m)
v1 = 6.4 m/s
In this problem, you were first asked to find how fast you
needed to throw the ball. In other words, you were asked to
find the initial velocity of the ball. Because all equations
include initial velocity, you could not use that as a criterion to
select the best formula. However, you know that the ball
reaches a maximum height (vy = 0 at the top of the motion) of
3.3 m, so the best relation to select is the one that relates
position and velocity. The speed of 6.4 m/s is about 14 mph,
so is a reasonable answer.
CHECK TO SEE IF REASONABLE AND WE
UNDERSTOOD PART B
y3 = y1 + v1Δt + ½ a(Δt)2
0 = 1.2 m + (6.4 m/s) Δt + ½ (-9.8 m/s2)( Δt)2
Δt = -0.16 s or +1.5 s
Δt = 1.5 s
In part b), you were asked to find how long (time) it took the
ball to reach the ground (position) and so the position-time
equation was used between the initial and final locations.
Because the position-time equation includes a ( Δt)2 term, it
will always give two answers for time. The solution that makes
physical sense is the positive time, or 1.5 s after the ball
leaves your hand.