Physics
Ms. Nguyen
Ch. 4- Projectile Problems
NAME: ________________________________________________
PERIOD: ________________
DATE: ___________________
Position-time equations for projectiles
Velocity-time equations for projectiles
(1)x f = xi + v x,i t
(3)v x, f = vx,i
1
(2) y f = yi + v y,i t - gt 2
2
(4)v y, f = v y,i - gt
Position-time equations for launch at angle
(5)x f = xi + (vi cosq )t
Velocity-time equations for launch at angle
(7)v x, f = vi cosq
1
(6) y f = yi + (vi sin q )t - gt 2
2
(8)v y, f = vi sin q - gt
Problem Type: Find final speed of a projectile when vertical displacement is not zero.
STEP 1: DAP
STEP 2: Find time in air.
STEP 3: Use time to find v y,f
*Remember: vx is constant
STEP 4: Find the final speed from vx,f and v y,f using Pythagorean theorem
1. A diver runs horizontally off the end of a diving board with an initial speed of 1.85 m/s. If the diving
board with is 3.00 m above the water, what is the diver’s speed just before she enters the water?
Strategy:
1) Find projectile time by using vertical position-time equation
2) Use time to find vertical final velocity
NB:
- Horizontal velocity will remain 1.85 m/s throughout the flight.
- The diver’s speed will be a combination of the vertical and horizontal
components of her velocity.
Solution: 1. Use the vertical positiontime equation to find the landing
time:
yf  yi  v y ,i t  12 g t 2
0  yi  0
1
2
2
 12 g tland
2
g tland
 yi
tland 

2  3.00 m
2 yi

 0.782 s
g
9.81 m s 2

2. Use the landing time to find v y , f :
v y , f  v y ,i  g t  0  9.81 m s 2  0.782 s   7.67 m/s
3. Find the final speed from vx , f and v y , f :
v  vx2, f  v y2, f 
1.85 m/s  2   7.89 m/s  2
 7.89 m/s
Insight: Projectile problems are often solved by first considering the vertical motion, which determines the
time of flight and the vertical speed, and then considering the horizontal motion.
YOUR TURN:
2. A flowerpot is knocked horizontally off a balcony with a speed of 3.5 m/s. If the balcony is 18.0 m above
the ground, what is the flowerpot’s final velocity?
Knowns:
yi = 18.0 m
yf = 0
vx,i = 3.5 m/s
vx,f = 3.5 m/s
vy,i = 0
Unknowns:
t
vy,f
vf
Notes:
- vx,i = vx,f
- to find t, use vy,i
Step 2: Find t
yf = yi + v y,it - g t
1
2
Step 3: Find vy,f
2
v = v x,f
+ v 2y,f =
v y,f = v y,i - g t
1
0 = 18.0m + 0t - (9.81m / s 2 )t 2
2
-18.0m = (-4.905m / s 2 )t 2
t=
Step 4: Find vf
2
)(
(
v y,f = 0 - 9.81 m s2 1.91565 s
)
v y,f = -18.79255 m/s
-18.0m
= 1.91565s
-4.905m / s 2
v=
(3.5 m/s)
2
(
+ -18.79255 m/s
)
2
v = 19.116m / s = 7.89 m/s
Problem Type: Find initial speed of a projectile launched horizontally
STEP 1: DAP
STEP 2: Find the time of flight
STEP 3: Use time to find initial speed
3. In Denver, after Halloween, children bring their jack-o’-lanterns to the top of a tower and compete for
accuracy in hitting a target on the ground. Suppose the tower is 9,0 m high and that the bull’s eye is a
horizontal distance of 3.5 m from the launch point. If the pumpkin is thrown horizontally, what is the
launch speed needed to hit the bull’s eye?
Strategy:
1)Find the time required for the pumpkin to drop 9.0 m,
2) Use that time to figure out the horizontal launch velocity (which will not
change throughout the flight) that will land the pumpkin on the target.
Solution: 1. Use the vertical
position-time equation to
find the time of flight:
yf  yi  v y ,i t  12 g t 2
yf  yi  0  12 g t 2
2  yf  yi 
g
2. Use the horizontal
position-time equation to
find the launch velocity:
t
2  0  9.0 m 

 9.81 m s 2

 1.355 s
xf = xi + v x,it
xf - xi
3.5- 0 m
= vx,i =
= 2.6 m/s
t
1.355 s
Insight: It is often the case that the vertical portion of a projectile’s motion is used to determine time
information about the trajectory, and the horizontal portion is used to find distance information.
YOUR TURN:
4. A beach ball is thrown horizontally from a lifeguard tower 3.2 m above the ground and lands 4.9 m from
the base of the lifeguard tower. What was the beach ball’s initial velocity if air resistance is neglected?
Knowns:
yi = 3.2 m
yf = 0
xi = 0
xf = 4.9 m
vx,i = 3.2 m/s
vx,f = 3.2 m/s
vy,i = 0
Unknowns:
t
vx,i
Step 2: Find t
yf = yi + v y,it - g t
1
2
Step 3: Find vx,i
2
1
0 = 3.2m + 0t - (9.81m / s 2 )t 2
2
-3.2m = (-4.905m / s 2 )t 2
t=
Notes:
- to find t, use vy,i
-3.2m
= 0.80771s
-4.905m / s 2
xf = xi + v x,i t
4.9m = 0 + v x,i (0.80771s)
v x,i = 6.066m / s = 6.1 m/s
Problem Type: Find x and y positions of a projectile at a certain time or find x and y velocity of a projectile at a
certain time
STEP 1 for position: DAP
STEP 2 for position: Use the position-time equations for projectiles launched at an angle θ in order to find the x
and y positions of the soccer ball at a given time.
STEP 1 for velocity: DAP
STEP 2 for velocity: Use the velocity-time equations for projectiles launched at an angle θ in order to find the x
and y components of the soccer ball’s velocity at a given time.
5. A soccer ball is kicked from the ground with an initial speed of 12 m/s at an angle of 32° above the
horizontal.
a. What are the x and y positions of the ball 0.50 s after it is kicked?
b. What are the x and y components of the football’s velocity 0.50 s after is it kicked?
41. Picture the Problem: A soccer ball is kicked from level ground
and travels along a parabolic trajectory.
Strategy: Use the position-time equations for projectiles launched
at an angle in order to find the x and y positions of the soccer ball
at a given time.
Solution: 1. The horizontal
velocity remains the same
throughout the flight:
xf  xi   vi cos   t
 0  12 m/s  cos 32  0.50 s 
xf  5.1 m
yf  yi   vi sin   t  12 g t 2
2. The vertical
position is affected by
the acceleration of
gravity:


 0  12 m/s  sin 32  0.50 s   12 9.81 m s 2  0.50 s   2.0 m
2
Insight: The ball will arrive at the peak of its flight at t  vi sin  g  0.648 s, so the ball has not yet reached
the peak. It’s peak height will be ymax  2.06 m when its downfield position is x = 6.60 m.
Strategy: Use the velocity-time equations for projectiles launched
at an angle in order to find the x and y components of the soccer
ball’s velocity at a given time.
Solution: 1. The horizontal
velocity remains the same
throughout the flight:
2. The vertical velocity is
affected by the acceleration of
gravity:
vx , f  vi cos 
 12 m/s  cos32  10 m/s
v y , f  vi sin   g t


 12 m/s  sin 32  9.81 m s 2  0.50 s   1.5 m/s
Insight: The ball will arrive at the peak of its flight at t  vi sin  g  0.648 s, so the ball has not yet reached
the peak. Its velocity at the peak will by 10 m/s in the horizontal direction, and its acceleration at that
instant will be 9.81 m/s2 in the downward direction, just as it is throughout the entire flight.
YOUR TURN:
6. A football is kicked from the ground with an initial speed of 14 m/s at an angle of 34° above the
horizontal.
a. What are the x and y positions of the ball 0.75 s after it is kicked?
b. What are the x and y components of the football’s velocity 0.75 s after is it kicked?
Knowns:
vi = 14 m/s
θ = 34°
t = 0.75 s
xi = 0
yi = 0
Unknowns:
xf
yf
a) Step 2: Find xf
xf = xi + ( vi cosq ) t
(
)
Notes:
Step 3: Find yf
yf = yi + ( vi sin q ) t - 12 g t 2
(
= 0 + éë 14 m/s cos34°ùû 0.75 s
)
(
)
(
xf = 8.705m = 8.7 m
yf = 3.1125m = 3.1 m
b) Step 2: Find vx,f
v x,f = vi cosq
Step 3: Find yy,f
v y,f = vi sin q - g t
(
)
= 14 m/s cos34°
v x,f = 11.60m / s = 12 m/s
) (
)(
= 0 + éë 14 m/s sin 34°ùû 0.75 s - 12 9.81 m s2 0.75 s
(
)
(
)(
= 14 m/s sin 34° - 9.81 m s2 0.75 s
v y,f = 0.4712m / s = 0.47 m/s
)
)
2
Problem Type: Find launch angle
STEP 1: DAP
STEP 2: Set vpeak  vx and solve for θ
7. A water balloon is launched in the air with an initial speed of 14 m/s. At its highest point its speed is 7.0
m/s. What was the launch angle of the projectile?
82. Picture the Problem: A projectile that is launched at an angle above horizontal follows a parabolic path.
Strategy: The projectile is accelerated only by gravity, so it maintains its horizontal velocity while its vertical velocity
is reduced from a large positive value (at launch) to zero (at the peak of its flight). Therefore the speed of the
projectile at the peak of its flight is equal to its horizontal speed at launch. Use this fact to determine the launch angle.
Solution: Set vpeak  vx and solve for :
vpeak  vx  vi cos 
æv ö
æ 7 m/s ö
x
÷÷ = cos-1 ç
÷ = 60°
è 14 m/s ø
è vi ø
q = cos-1 çç
Insight: If the launch angle were 45° the speed at the peak would be vi cos 45  12 m/s cos 45  8.5 m/s.
YOUR TURN:
8. A bullet is launched in the air with an initial speed of 98 m/s. At its highest point its speed is 49 m/s.
What was the launch angle of the bullet?
Knowns:
vi = 98 m/s
vi = 49 m/s
Unknowns:
xf
Step 2: Set vpeak  vx and solve for 
vpeak  vx  vi cos 
Notes:

æv ö
æ 49 m/s ö
x
÷÷ = cos-1 ç
÷ = 60°
è 98 m/s ø
è vi ø
q = cos-1 çç
Note:
- Only vertical velocity will determine how long a projectile is in the air
- To determine how far a projectile travels (horizontal displacement), multiply time by horizontal velocity
-