Copyright © 2011 Nelson Education Ltd. Chapter 2: Motion in Two Dimensions 2.3-1

Section 2.3: Projectile Motion

Tutorial 1 Practice, page 78 1. (a) Given: ∆dy = –32 m; ay = –9.8 m/s

2;

vy = 0 m/s

Required: ∆t

Analysis:

!dy

= vy! t +

1

2a

y! t2

!dy

= 0 +1

2a

y! t2

! t2=

2!dy

ay

! t =

2!dy

ay

Solution:

! t =

2!dy

ay

=

2 !32 m( )

!9.8 m

s2

"#$

%&'

= 2.556 s

! t = 2.6 s

Statement: The hockey puck is in flight for 2.6 s.

(b) Given: ∆t = 2.6 s; ax = 0 m/s2; vx = 8.6 m/s

Required: ∆dx

Analysis: !d

x= v

x! t

Solution:

!dx

= vx! t

= 8.6 m

s

!"#

$%&

2.556 s( ) (two extra digits carried)

!dx

= 22 m

Statement: The range of the hockey puck is 22 m.

2. Since the velocity of the puck does not affect its

vertical motion, it would still take 2.6 s to hit the

ground. The range of the puck would be half

because it is travelling at half the velocity

(horizontally). That means the range would be

11 m.

Tutorial 2 Practice, page 81 1. Given: ∆dy = –17 m; ay = –9.8 m/s

2;

vi = 7.3 m/s; θ = 25°

First determine the time of flight:

Required: ∆t

Analysis: !d

y= v

y! t +

1

2a

y! t2

Solution:

!dy

= vy! t +

1

2a

y! t2

= vi

sin!( )! t +1

2a

y! t2

"17 = 7.3( ) sin25°( )! t +1

2"9.8( )! t2

0 = 3.085! t " 4.9! t2+ 17

! t =

"3.085 ± 3.0852" 4 "4.9( ) 17( )

2 "4.9( )

="3.085 ± 18.51

"9.8

! t = "1.574 s or ! t = 2.204 s

The answer must be the positive value.

Statement: The superhero is in flight for 2.2 s.

Determine the range:

Required: ∆dx

Analysis: !d

x= v

x! t

Solution:

!dx

= vx! t

= vicos! ! t

= 7.3 m

s

"#$

%&'

cos25°( ) 2.204 s( ) (two extra digits carried)

!dx

= 15 m

Statement: The superhero travels 15 m

horizontally before landing.

Determine the final velocity:

Required:

!v

f

Analysis:

!v

f=!v

fx+!v

fy

Solution:

!v

f=!v

fx+!v

fy

= vicos!( ) +

!v

iy+!a

y! t( )

= vicos! + v

isin! +

!a

y! t

= 7.3 m

s

"#$

%&'

cos25°( ) [right] + 7.3 m

s

"#$

%&'

sin25°( ) [up]

+ 9.8 m

s2

[down]"#$

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2.204 s( ) (two extra digits carried)

= 6.616 m/s [right] + 3.085 [up] + 21.60 m/s [down]

= 6.616 m/s [right]( 3.085 [down] + 21.60 m/s [down]!v

f= 6.616 m/s [right] + 18.52 m/s [down]

Use the Pythagorean theorem:

vf

2= v

fx

2+ v

fy

2

vf

= vfx

2+ v

fy

2

= 6.616 m/s( )2

+ 18.52 m/s( )2

(two extra digits carried)

vf

= 2.0 ! 10 m/s

Copyright © 2011 Nelson Education Ltd. Chapter 2: Motion in Two Dimensions 2.3-2

Let φ represent the angle

!v

f makes with the x-axis.

tan! =

vfy

vfx

=

18.52 m

s

6.616 m

s

(two extra digits carried)

= 2.799

! = tan"1

2.799( )! = 70°

Statement: The superhero’s final velocity is

2.0 × 10 m/s [right 70° down].

2. The ball thrown at an angle above the horizontal

will take longer to reach the ground. It has an

initial velocity with a vertical component, so it will

take more time for it to reach the ground due to

gravity. The difference in the times to reach the

ground depends on the initial height and the initial

velocity of the first ball.

Section 2.3 Questions, page 81 1. The horizontal and vertical motions of a

projectile take the same amount of time.

2. Given: ∆dx = 20.0 m; ay = –9.8 m/s2;

vx = 10.0 m/s; vy = 0 m/s

Determine the time of flight first:

Required: ∆t

Analysis:

vx

=!d

x

! t

! t =!d

x

vx

Solution:

! t =!d

x

vx

=20.0 m

10.0 m

s

! t = 2.00 s

Statement: It takes the tennis ball 2.00 s to reach

the ground.

Determine the vertical displacement:

Required: ∆dy

Analysis: !d

y= v

y! t +

1

2a

y! t2

Solution:

!dy

= vy! t +

1

2a

y! t2

= 0 +1

29.8

m

s2

!

"#

$

%& 2.00 s( )

2

!dy

= 2.0 ' 10 m

Statement: The water tower is 2.0 × 10 m high. 3. (a) To give a projectile the greatest time of

flight, launch it at 90° from the ground because

this angle maximizes the vertical component of the

velocity. At 90° from the ground, all the velocity is

straight up instead of some of the velocity going

into the horizontal component.

(b) To give a projectile the greatest time of flight,

launch it at 45°. If the angle is less than 45°, the

flight will be too short to travel any farther. If the

angle is greater than 45°, the horizontal component

of the velocity is too short to travel any farther.

4. (a) The ball experiences no horizontal

acceleration. The ball accelerates 9.8 m/s2 down

due to gravity.

(b) The ball experiences no horizontal

acceleration. The ball accelerates 9.8 m/s2 down

due to gravity.

(c) The ball experiences no horizontal acceleration.

The ball accelerates 9.8 m/s2 down due to gravity.

5. The arrow strikes the ground before reaching the

target. It would take the arrow more than 1 s to

travel to 60 m to the target when travelling at

55 m/s. But in 1 s, the arrow would fall more than

1.5 m due to gravity. For her next shot, the archer

should increase the initial velocity, aim higher, or a

combination of the two.

6. (a) Given: vi = 26 m/s; θ = 60°; ay = –9.8 m/s2;

vfy = 0 m/s

Required: ∆t

Analysis:

vfy

= viy

+ ay! t

! t =

vfy! v

iy

ay

Solution:

! t =

vfy! v

iy

ay

=

0 ! 26 m/s( ) sin60°( )

!9.8 m/s2

=

22.52 m

s

9.8 m

s2

= 2.298 s

! t = 2.3 s

Statement: It takes the acrobat 2.3 s to reach his

maximum height.

(b) Given: vi = 26 m/s; θ = 60°; ay = –9.8 m/s2;

vfy = 0 m/s

Determine the maximum height:

Required: ∆dy

Analysis: !d

y=

vfy

+ viy

2! t

Copyright © 2011 Nelson Education Ltd. Chapter 2: Motion in Two Dimensions 2.3-3

Solution:

!dy

=

vfy

+ viy

2! t

=

0 ! 22.52 m

s

22.298 s( ) (two extra digits carried)

= 25.88 m

!dy

= 26 m

Statement: The maximum height of the acrobat is

26 m.

Determine the time to reach half the maximum

height (13 m) the second time:

Required: ∆t

Analysis: !d

y= v

iy! t +

1

2a

y! t2

Solution:

!dy

= viy! t +

1

2a

y! t2

13 = 22.52( )! t +1

2!9.8( )! t2

0 = 22.52! t ! 4.9! t2! 13

! t =

!22.52 ± 22.522! 4 !4.9( ) !13( )

2 !4.9( )

=!22.52 ± 15.89

!9.8

! t = 0.68 s or ! t = 3.9 s

The first time is on his way up, so the correct time

is 3.9 s.

Statement: It takes the acrobat 3.9 s to reach a

point halfway back down to the ground.

7. Given: vi = 20 m/s; θ = 45°; ay = –9.8 m/s2;

vfy = 0 m/s

Determine the time of flight:

Required: ∆t

Analysis: !d

y= v

iy! t +

1

2a

y! t2

Solution:

!dy

= viy! t +

1

2a

y! t2

0 = vi

sin45°( )! t +1

2a

y! t2

0 = vi

sin45°( ) +1

2a

y! t, ! t ! 0

1

2a

y! t = "v

isin45°( )

! t =

"2vi

sin45°( )

ay

=

!2 20 m

s

"#$

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sin45°( )

!9.8 m

s2

"#$

%&'

= 2.886 s

! t = 2.9 s

Statement: The time of flight of the golf ball is

2.9 s.

Determine the horizontal distance:

Required: ∆dx

Analysis: !d

x= v

ix! t

Solution:

!dx

= vix! t

= vi

cos45°( )! t

= 20 m

s

!"#

$%&

sin45°( ) 2.9 s( )

!dx

= 41 m

Statement: The golfer was 41 m from the hole

when he hit the ball.

Determine the maximum height:

Required: ∆dy

Analysis:

vfy

2= v

iy

2+ 2a

y!d

y

!dy

=

vfy

2! v

iy

2

2ay

Solution:

!dy

=

vfy

2 ! viy

2

2ay

=

0 ! visin45°( )

2

2 !9.8 m/s2

( )

=

0 ! 20 m/s( ) sin45°( )"#

$%

2

!19.6 m/s2

=

200 m

2

s2

19.6 m

s2

!dy

= 1.0 & 10 m

Statement: The golf ball reached a maximum

height of 1.0 × 10 m.

8. Given: ∆dy = –12 m; ay = –9.8 m/s2;

vi = 4.5 m/s; θ = 25°

First determine the time of flight:

Required: ∆t

Analysis: !d

y= v

y! t +

1

2a

y! t2

Copyright © 2011 Nelson Education Ltd. Chapter 2: Motion in Two Dimensions 2.3-4

Solution:

!dy

= vy! t +

1

2a

y! t2

= vi

sin!( )! t +1

2a

y! t2

"12 = 4.5( ) sin25°( )! t +1

2"9.8( )! t2

0 = 1.902! t " 4.9! t2+ 12

! t =

"1.902 ± 1.9022" 4 "4.9( ) 12( )

2 "4.9( )

="1.902 ± 15.45

"9.8

! t = "1.382 s or ! t = 1.771 s

The answer must be the positive value.

Statement: The beanbag is in flight for 1.8 s.

Determine the range:

Required: ∆dx

Analysis: !d

x= v

x! t

Solution:

!dx

= vx! t

= vicos! ! t

= 4.5 m

s

"#$

%&'

cos25°( ) 1.771 s( ) (two extra digits carried)

!dx

= 7.2 m

Statement: The student’s friend must stand 7.2 m

from the building to catch the beanbag.