1. Example Problems for Motion in 2-d Answers
1. A stone is thrown horizontally at 15 m/s from a cliff
44 m high.
(a) How long does it take the stone to reach the ground?
vx = 15 m/s
Stone takes parabolic path to the ground
Gravity acts in a vertical direction; horizontal velocity as no effect on gravity
44 m B A
So gravity acts equally on object projected horizontally (which
lands at A), and object that is dropped (which lands at B)
To find time of flight to A (which is difficult), find instead the
time it takes to drop the same vertical distance to B (easier)

2. t = 3.0 s (a) t = ? Vertically, vi = 0 g = - 9.8 m/s2 d = - 44 m
vx = 15 m/s
g = m/s d = m
Use d = vi t + ½ g t2
44 m
d = ½ g t2 B A
2 (
d = ½ g t2
) 2
2 ( - 44 m )
2 d = g t2
2 d
t = =
- 9.8 m/s2 g g g
2 d
t2 = g
t = 3.0 s

3. (b) How far from the base of the cliff does it land?
vx = 15 m/s
t = 3.0 s
Gravity acts vertically;
horizontal motion is unaffected
44 m
dx = ?
Motion in horizontal is at constant velocity
dx = vx t
= ( 15 m/s )( 3.0 s )
dx = 45 m

4. 2. A stone is thrown horizontally at 10.0 m/s from a cliff
78.4 m high.
(a) How long does it take the stone to reach the ground?
vx = 10.0 m/s
Instead of finding time of flight to A,
find the time it takes to drop the
same vertical distance to B
78.4 m B A

5. t = 4.0 s (a) t = ? Vertically, vi = 0 g = - 9.8 m/s2 d = - 78.4 m
vx = 10.0 m/s
g = m/s d = m
Use d = vi t + ½ g t2
78.4 m
d = ½ g t2 B A
2 (
d = ½ g t2
) 2
2 ( m )
2 d = g t2
2 d
t = =
- 9.8 m/s2 g g g
2 d
t2 =
t = 4.0 s g

6. (b) How far from the base of the cliff does it land?
vx = 10.0 m/s
t = 4.0 s
Gravity acts vertically;
horizontal motion is unaffected
78.4 m
dx = ?
Motion in horizontal is at constant velocity
dx = vx t
= ( 10.0 m/s )( 4.0 s )
dx = m

7. 3. vx = 0.800 m/s Height of table = 0.960 m dx = vx t t = ?
Use dy = viy t + ½ g t2
dx = ?
2 dy
see earlier examples
t = g

8. Time of flight for object dropped from height h
3. vx = m/s
Height of table = m
vx = m/s
dx = vx t
t = ?
0.950 m
Use dy = viy t + ½ g t2
dx = ?
Time of flight for object dropped from height h
2 dy
2 h
t =
t = g g

9. 3. vx = 0.800 m/s h = 0.960 m dx = vx t t = ? 2 h t = g 2 ( 0.950 m )
= t = s =
9.8 m/s2
Then dx = vx t
= ( m/s )( s )
dx = m

10. 4. dx = 52 m h = 122 m (a) Time of flight = ? 2 h t = g 2 ( 122 m ) =
9.8 m/s2
t = 5.0 s

11. 4. dx = 52 m h = 122 m (b) vx = ? dx = vx t t t dx 122 m vx = = t
t = 5.0 s
dx = vx t
52 m t t dx
122 m
vx = = t
5.0 s
vx = 24 m/s

12. 5. t = 4.5 s dx = 36 m (a) vx = ? dx = vx t t t dx 36 m vx = = t 4.5 s
vx = 8.0 m/s

13. 5. t = 4.5 s dx = 36 m (b) dy = ? Vertically, vi = 0
vx = 8.0 m/s
(b) dy = ?
dy = ?
t = 4.5 s
Vertically, vi = 0
dy = vi t + ½ g t2
36 m
= ½ g t2
= ½ g t2
= ½ ( m/s2 )( 4.5 s )2
dy = m
So cliff is 99 m high

14. 6. A projectile is fired with a velocity of 196 m/s at an
angle of 60.0o with the horizontal.
(a) Find the vertical and horizontal velocities
Draw horizontal and
vertical components
196 m/s vy
Use SOH-CAH-TOA
60o vx

15. SOH-CAH-TOA vy vx vx : vy : adj opp cos 60 = sin 60 = hyp hyp vx 196
196 m/s vy
60o vx
vx :
vy :
adj
opp
cos 60 =
sin 60 =
hyp
hyp vx
196 vy
196
(196) =
(196) =
196
196
vx = 98 m/s
vy = 170 m/s

16. (b) Find the time the projectile is in the air
Projectile takes parabolic
path to the ground
170 m/s
As in earlier case, gravity is
unaffected by
horizontal motion
98 m/s
Time of flight depends only on initial vertical velocity

17. (b) Find the time the projectile is in the air
Projectile takes parabolic
path to the ground
170 m/s
As in earlier case, gravity is
unaffected by
horizontal motion
98 m/s
Time of flight depends only on initial vertical velocity
Time of flight will be the same for this projectile and
for a projectile shot straight up with an initial velocity
of 170 m/s

18. As in earlier cases, consider only 1st half of flight : vf = 0
vi = 170 m/s g = m/s2
vf = 0
t = ?
As in earlier cases, consider only
1st half of flight
170 m/s
: vf = 0
vf = vi + g t
98 m/s
0 = vi + g t
- vi
- vi
- 170 m/s
t =
= s
- vi = g t
- 9.8 m/s2 g g
Multiply by 2 to get
total flight time:
- vi
t =
tt = 2 ( 17.3 s ) = tt = s g

19. (c) Find the horizontal distance the projectile travels (this
is called the range)
Again, horizontal motion is
affected only by
horizontal velocity
t = s
170 m/s
98 m/s
dx = ?
dx = vx t
= ( 98 m/s )( 34.6 s )
dx = m

20. (d) Find the maximum height the projectile reaches
Height reached depends only on
initial vertical velocity and time to reach
the highest point
170 m/s
vi = 170 m/s g = m/s2
t = s (time of 1st half)
98 m/s
dy = vi t + ½ g t2
= ( 170 m/s )( 17.3 s ) + ½ ( m/s2 )( 17.3 s )2
= m + ( m )
dy = m

21. 7. A projectile is fired at an angle of 53.0o with the
horizontal at a velocity of 200 m/s.
(a) Find the time the
projectile is in the air
200 m/s vy
Use SOH-CAH-TOA to find
components of velocities
53o vx
vx :
vy :
adj
opp
cos 53 =
sin 53 =
hyp
hyp vx vy
200
200
(200) =
(200) =
200
200
vx = 120 m/s
vy = 160 m/s

22. Use vertical velocity to find time of flight vf = 0
vi = 160 m/s g = m/s2
t = ?
vf = ( for 1st half )
160 m/s
vf = vi + g t
120 m/s
0 = vi + g t
- vi
- vi
- 160 m/s
t =
= s
- vi = g t
- 9.8 m/s2 g g
Multiply by 2 to get
total flight time:
- vi
t =
tt = 2 ( 16.3 s ) = tt = s g

23. (b) Find the horizontal distance the projectile travels (the range)
t = 32.6 s
160 m/s
120 m/s
dx = ?
dx = vx t
= ( 120 m/s )( 32.6 s )
dx = m

24. 8. A golf ball is hit at an angle of 45o with the
horizontal at a velocity of 52 m/s. Find the range.
52 m/s
45o
dx = ?
dx = vx t
Need to find vx and
the time of flight
First, find the velocity components
using SOH-CAH-TOA

25. vy vx vx : vy : adj opp cos 45 = sin 45 = hyp hyp vx vy 52 52 (52)
52 m/s vy
45o vx
vx :
vy :
adj
opp
cos 45 =
sin 45 =
hyp
hyp vx vy 52 52
(52) =
(52) = 52 52
vx = m/s
vy = m/s

26. Use vertical velocity to find time of flight vf = 0
vi = m/s g = m/s2
t = ?
vf = ( for 1st half )
36.8 m/s
vf = vi + g t
36.8 m/s
0 = vi + g t
- vi
- vi
m/s
t =
= s
- vi = g t
- 9.8 m/s2 g g
Multiply by 2 to get
total flight time:
- vi
t =
tt = 2 ( 3.75 s ) = tt = s g

27. dx = vx t = ( 36.8 m/s )( 7.50 s ) dx = 276 m dx = ? t = 7.50 s