# Kinematics II A/S Physics

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14 Νοε 2013 (πριν από 4 χρόνια και 7 μήνες)

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Kinematics II

A/S Physics

Use
g=
9.81ms
-
2
.

Answers

1.

A particle is projected with a speed of 21ms
-
1

at an angle of elevation of
60 degrees. Find its speed and direction of motion after (a) 1 second (b) 2
seconds (c) 3 second
s.

2.

A ball is thrown from O with a speed of 28ms
-
1

at an angle of elevation of
60 degrees. It hits a wall which is 5m horizontally from O. Calculate the
height above O at which the ball hits the wall.

3.

A ball is thrown at 14ms
-
1

at an angle of elevat
ion of 60 degrees. Find its
horizontal and vertical distances from the point of projection after 1
second. Find the direction of motion of the ball at this time.

4.

A ball is thrown from O with a speed of 30ms
-
1

at an angle of 30 degrees
above the horizon
tal.
Given that it just clears the top of a wall at a
horizontal distance of 20m from O, find the height of the top of the wall
above O.

5.

A gun fires a shell with an initial velocity of 770ms
-
1

at an angle of 18
degrees above the horizontal. By modellin
g the shell as a particle find the
range of the shell.

6.

A stone is thrown at an angle of elevation of 30 degrees. If 1 second later
it hits the ground 1m below its point of projection find the speed of
projection. Find its greatest height above the poin
t of projection.

Kinematics II
-

Answers

A/S Physics

1.

A particle is projected with a speed of 21ms
-
1

at an angle of elevation of
60 degrees. Find its speed and direction of motion after (a) 1 second (b) 2
seconds (c) 3 seconds.

Answer

The horizontal

component of
velocity does not change
, it remains

5
.
10
60
cos
21

x
u
ms
-
1
. The vertical velocity is affected by gravity.
Initially it is
60
sin
21

y
u
, so that the vertical
component of velocity is
given by
at
u
v
y
y

.

Plugging
in the numbers:

1
-
1
-
1
-
ms
2
.
11
3
ms
43
.
1
2
ms
38
.
8
1
81
.
9
60
sin
21

y
y
y
y
v
v
v
t
t
v

1
2
2
1
2
2
1
2
2
2
2
ms
4
.
15
2
.
11
5
.
10
)
3
(
ms
6
.
10
43
.
1
5
.
10
)
2
(
ms
43
.
13
38
.
8
5
.
10
)
1
(
)
(

v
v
v
v
v
t
v
y
x

The direction of motion is given by

x
y
x
y
v
v
v
v
1
tan
tan

So plugging in the numbers again (the angle is a function of time):

8
.
46
5
.
10
2
.
11
tan
3
76
.
7
5
.
10
43
.
1
tan
2
6
.
38
5
.
10
38
.
8
tan
1
tan
1
1
1
1

x
y
v
v
t

The direction of motion is alwa
ys the same as the direction of the velocity
vector.

2.

A ball is thrown from O with a speed of 28ms
-
1

at an angle of elevation of
60 degrees. It hits a wall which is 5m horizontally from O. Calculate the
height above O at which the ball hits the wall.

An
swer
:

Calculate the time of flight first, using the horizontal distance and
velocity:

s
357
.
0
60
cos
28
5

t

Now, the height after that time:

m
03
.
8
60
cos
28
5
2
81
.
9
60
cos
28
5
60
sin
28
2
2
2
1

at
ut
s

3.

A ball is thrown at 14ms
-
1

at an angle of elevation of 60 degrees. Find its
ho
rizontal and vertical distances from the point of projection after 1
second. Find the direction of motion of the ball at this time.

Answer
:

Horizontally:
60
cos
14

u

and remains constant so distance covered after
one second is 7m

Vertically
its distance is given by:

m
22
.
7
1
81
.
9
60
sin
14
2
2
1

s
t
t
t
s

Its direction of motion is the same as the direction of the velocity vector.
After 1 second the horizontal component is still 7ms
-
1

and the vertical
component is:

1
ms
31
.
2
1
81
.
9
60
sin
14
1

v
at
u
v

Hence the dire
ction of motion is

3
.
18
7
81
.
9
60
sin
14
tan
7
81
.
9
60
sin
14
tan
1

4.

A ball is thrown from O with a speed of 30ms
-
1

at an angle of 30 degrees
above the horizontal. Given that it just clears the top of a wall at a
horizontal distance of 20m from O, find the height of the top o
f the wall
above O.

Answer

Just need to work out the vertical displacement after covering 20m. So
use time in flight whilst covering the horizontal distance of 20m and then
use that time to determine vertical displacement:

Time in flight is:

s
770
.
0
30
cos
30
20

t

And vertical displacement:

64
.
8
30
cos
30
20
2
81
.
9
30
cos
30
20
30
sin
30
2
2
2
1

at
ut
s

5.

A gun fires a shell with an initial velocity of 770ms
-
1

at an angle of 18
degrees above the horizontal. By modelling the shell as a particle find the
range of the shell.

Answer
:

Use the fact that the vertical displacement will be zero at t=0 and when it
next hits the ground (assuming the ground is horizontal!), so:

0
0
0
0
2
1
2
1
2
2
1
at
u
t
at
u
t
at
ut

Using the last equation:

s
5
.
48
81
.
9
18
sin
770
2
0
81
.
9
18
sin
770
0
2
1
2
1

t
t
at
u

Now use the horizontal component o
f velocity to determine the range:

m
500
,
35
81
.
9
18
sin
770
2
18
cos
770

s

6.

A stone is thrown at an angle of elevation of 30 degrees. If 1 second later
it hits the ground 1m below its point of projection find the speed of
projection. Find its greatest height above th
e point of projection.

Answer
:

Consider the vertical displacement first. Using

2
2
1
at
ut
t
s

, we have:

1
ms
905
.
3
2
81
.
9
1
1
1

y
y
u
u
s

Now we can just use trig to find its
speed

of projection:

1
ms
81
.
7
30
sin
30
sin

y
y
u
u
u
u

Its greatest height:

m
777
.
0
81
.
9
2
905
.
3
81
.
9
2
905
.
3
0
2
2
2
2
2

s
s
as
u
v