Note Chap 2 SP015 (Kinematics of linear motion) Zahasnida

Chapter 02 Physics

2.3.4 Horizontal range, R and value of R maximum

 Since the x-component for velocity along AC is constant hence

ux = vx = u cos

 From: s x = u xt and s x = R

R = (u cos  )(t )

R = (uu2c(o2ssin)2cuosgsin  
R =
g )

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Chapter 02 Physics

 From : sin 2 = 2 sin  c o s 

R = u 2 sin 2

g

 The R is maximum when  = 45 and sin 2 = 1

Rmax = u2
g

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Chapter 02 Physics

Example 2.7 :

A tennis ball is thrown upward from the top of a building with
velocity 15 m s-1 at an angle 30 to the horizontal. The height of
the building is 40 m. Calculate

a. the maximum height of the ball from the ground.

b. the magnitude of the velocity of the ball just before it strikes the

ground. (given g = 9.81 m s-2) ux = u cos 30
Solution : u = 15 m s−1 = 15 cos 30
= 13.0 m s−1
30

ax = 0 H =? uy = u sin 30
ay = −g = 15 sin 30
h = 40 m = 7.50 m s−1

v = ? 53

Chapter 02 Physics

Solution : sy u v v x = v
a. v y = 0
30 s H =?
sx

h = 40 m

vy2 = uy2 + 2aysy 54

0 = (7.50)2 + 2 (−9.81) sy

sy = 2.87 m

H = sy + h
= 2.87 + 40

H = 42.9 m

Chapter 02 Physics

Solution : u
b.
30 sx

h = 40 m sy = −40 m s

vx = u x= 13.0 m s − 1 vy vx
v = uy 2+ v=?
2 .50 )22a+y sy ( −
y = (7 2 ) ( ) 55

9 .8 1 − 4 0

vy = 29.0 m s−1
v = vx2 + vy2

= (13.0)2 + ( −29.0)2
v = 31.8 m s−1

Chapter 02 Physics

2.3.5 Horizontal projectile

 A ball bearing rolling off the end of a table with an initial

velocity, u in the horizontal direction. sx
uu

vx

h sy  vy v
s

Figure 2.6 A x B

 Horizontal component : v elo c ity , u x = u = v x = c o n s tan t 56

displacement, sx = x

 Vertical component :

initial velocity, uy = 0

displacement, sy = −h

Chapter 02 Physics

Time taken for the ball to reach the floor (point B), t

sy = uyt − 1 gt 2
2
−h = 0− 1
gt 2
2

t = 2h
g

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Chapter 02 Physics

Horizontal displacement, x

 Use condition: The time taken for the = The time taken for the
ball free fall to point A ball to reach point B

s x = u x t and s x = x

x = u 2h 
g

 Note : Figure 2.7

⚫ In solving any calculation problem about projectile motion,

the air resistance is negligible.

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Chapter 02 Physics

Example 2.8 :
y

H
u

Figure 2.8 O  = 60.0 P v1x x
v2x
R v1
Figure 2.8 shows a ball thrown by superman with v1y v2
an initial speed, u = 200 m s-1 and makes an angle, Q

 = 60.0 to the horizontal. Determine v2y

a. the position of the ball, and the magnitude and

direction of its velocity, when t = 2.0 s.

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Chapter 02 Physics

b. the time taken for the ball reaches the maximum height, H and
calculate the value of H.

c. the horizontal range, R
d. the magnitude and direction of its velocity when the ball

reaches the ground (point P).
e. the position of the ball, and the magnitude and direction of its

velocity at point Q if the ball was hit from a flat-topped hill with
the time at point Q is 45.0 s.
(given g = 9.81 m s-2)
Solution :
Component of Initial velocity :

ux = 200 cos 60.0 = 100 m s−1
uy = 200 sin 60.0 = 173 m s−1

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Chapter 02 Physics

Solution :
a. i. position of the ball when t = 2.0 s ,

Horizontal component :

sx = uxt

s x = (100 )(2.00 )

sx = 200 m from point O

Vertical component : sy = uyt − 1 gt2
2
1
sy = (173)(2.00) − 2 (9.81)(2.00 )2

sy = 326 m above the ground

therefore the position of the ball is (200 m, 326 m)

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Chapter 02 Physics

Solution :

a. ii. magnitude and direction of ball’s velocity at t = 2.0 s ,

Horizontal component : v x = u x = 1 0 0 m s −1

Vertical component : v y = u y − g t

= (173) − (9.81)(2.00 )

vy = 153 m s−1

Magnitude, v = v 2 + v 2 = (100 )2 + (153 )2
x y

v = 183 m s−1

Direction,  vy  = tan −1 153  y
vx  100 
θ = tan −1 
v 56.8
θ = 56.8
x

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Chapter 02 Physics

Solution :

b. i. At the maximum height, H : v y = 0

vy = uy − gt

0 = (173 ) − (9.81)t

t = 17.6 s

ii. 1 2
2
sy = u yt − gt

H = (173)(17.6) − 1 (9.81)(17.6)2
2
H = 1525 m

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Chapter 02 Physics

Solution :

c. Flight time = 2(the time taken to reach the maximum height)

t = 2(17.6 )

t = 35.2 s

Horizontal range, R is

sx = uxt

R = (100 )(35.2 )

R = 3520 m

d. When the ball reaches point P thus s y = 0

The velocity of the ball at point P,

Horizontal component: v1x = u x = 1 0 0 m s −1
= u(1y7−3
Vertical component: v1 y = gt

)− (9.81)(35.2 )
v1y = −172 m s −1
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Chapter 02 Physics

Solution :

( ) ( )Magnitude, v1 = v12x + v12y = 100 2 + − 172 2

v1 = 200 m s −1

Direction, tan −1  v1 y  = tan −1  −172  y
v1x  100 
θ=

θ = −60.0  60.0 x

v1 65

e. The time taken from point O to Q is 45.0 s.

i. position of the ball when t = 45.0 s,

Horizontal component : s x = u x t

s x = (100 )(45.0 )

sx = 4500 m from point O

Chapter 02 Physics

Solution :

Vertical component : 1 gt2
2
sy = u yt −

s y = (173)(45.0) − 1 (9.81)(45.0 )2
2
sy = −2148 m below the ground

therefore the position of the ball is (4500 m, −2148 m)

e. ii. magnitude and direction of ball’s velocity at t = 45.0 s ,

Horizontal component : v 2 x = u x = 1 0 0 m s −1

Vertical component : v 2 y = u y − g t

= (173) − (9.81)(45.0 )

v2 y = −269 m s−1

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Chapter 02 Physics

Solution :

Magnitude, v2 = v 2 + v 2 y
2x 2

v2 = (1 0 0 )2 + (− 2 6 9 )2
v2 =
287 m s −1

Direction, θ = tan −1  v2 y 
v2 x

θ = tan −1 − 269  y x
 100 
 69.6
θ = −69.6 v2

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Chapter 02 Physics

Example 2.9 :

A transport plane travelling horizontally at a constant velocity of
50 m s−1 at an altitude of 300 m releases a parcel when directly
above a point X on level ground. Calculate

a. the flight time of the parcel,

b. the velocity of impact of the parcel,

c. the distance from X to the point of impact.

(given g = 9.81 m s-2)

Solution : u = 50 m s−1

300 m 68
X

d

Chapter 02 Physics

Solution :

The parcel’s velocity = plane’s velocity

u = 50 m s−1
thus u x = u = 50 m s −1 and u y = 0 m s −1

a. The vertical displacement is given by

sy = −300 m

Thus the flight time of the parcel is
1 gt2
− sy = u yt − 2 2
300 = 0 − 1
(9.81)t
2
t = 7.82 s

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Chapter 02 Physics

Solution :

b. The components of velocity of impact of the parcel:

Horizontal component: v x = u x = 50 m s −1

Vertical component: v y = u y − g t

= 0 − (9.81)(7.82 )

vy = −76.7 m s−1

Magnitude, v= v 2 + v 2 = (50)2 + (− 76.7 )2
x y

v = 91.6 m s−1

y

Direction, t−a5n6−.19vv y  = tan −1  − 76.7   x
x  50  v2
θ= 56.9

θ= 70

Chapter 02 Physics

Solution :
c. Let the distance from X to the point of impact is d.

Thus the distance, d is given by

sx = uxt

d = (50)(7.82)

d = 391m

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Chapter 02 Physics

Exercise 2.3 :

Use gravitational acceleration, g = 9.81 m s−2
1. A basketball player who is 2.00 m tall is standing on the floor

10.0 m from the basket, as in Figure 2.13. If he shoots the
ball at a 40.0 angle above the horizontal, at what initial
speed must he throw so that it goes through the hoop without
striking the backboard? The basket height is 3.05 m.

Figure 2.9 72

ANS. : 10.7 m s−1

Chapter 02 Physics

2. An apple is thrown at an angle of 30 above the horizontal
from the top of a building 20 m high. Its initial speed is
40 m s−1. Calculate
a. the time taken for the apple to strikes the ground,
b. the distance from the foot of the building will it strikes
the ground,
c. the maximum height reached by the apple from the
ground.

ANS. : 4.90 s; 170 m; 40.4 m

3. A stone is thrown from the top of one building toward a tall
building 50 m away. The initial velocity of the ball is 20 m s−1
at 40 above the horizontal. How far above or below its
original level will the stone strike the opposite wall?

ANS. : 10.3 m below the original level.

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Chapter 02 Physics

THE END.

Next Chapter…

CHAPTER 3 :
Momentum and Impulse

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