ONLINE: MATHEMATICS EXTENSION 2
Topic 6
MECHANICS
EXERCISE p6201
Two stones are thrown simultaneously from the same point in the same upward
direction at the angle  with respect to the horizontal. The initial velocities of the two
stones are u1 and u2 where u1 < u2.The slower stone hits the ground at a point P on the
same level as the point of projection. At that instant the faster stone just clears a wall of
height h above the level of projection and its downward path makes an angle  with the
horizontal.
(a)
Show that the line joining them has an inclination with respect to the horizontal
that is independent of time when the stones are still both in flight.
(b)
What is the horizontal displacement from the point P to the foot of the wall in
terms of  and h?
(c)
Show that
(d)
If   2  show that
u2  tan   tan    2 u1 tan 
u1  43 u2
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Solution
Step 1: Think about how to approach the problem
Step 2: Draw an annotated diagram of the physical situation
Step 3: What do you know about projectile motion and motion with a constant acceleration?
The origin is at 0(0, 0).
Stone 1 has initial velocity u1
u1 x  u1 cos 
u1 y  u1 sin 
tan    u1 y / u1 y 
audio
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Stone 2 has initial velocity u2
u2 x  u2 cos 
u2 > u1
u2 y  u1 sin 
tan    u2 y / u2 y 
Stone 1 lands at the point P(xP, 0). At this instance stone 2 is at the point Q(xQ, h).
The velocity of stone 2 at the point Q is vQ
vQx  vQ cos 
uQy  vQ sin 
tan    vQy / vQx 
(a)
When stone 1 is at the point S(x1, y1) then stone 2 is at the point T(x2, y2).
The inclination of the line joining the stones 1 and 2 makes an angle  with the
horizontal where
tan  
y2  y1
x2  x1
Therefore, we need to find the coordinates of the points S(x1, y1) and T(x2, y2) at some
instance t.
a1 x  0 a1 y   g
v1 y  0
Motion with constant acceleration
v  u  at
physics.usyd.edu.au/teach_res/hsp/math/math.htm
v 2  u 2  2a s
p6201
s  u t  12 a t 2
3
x1  u1x t   u1 cos  t
x2   u2 cos   t
y1  u1 y t  12   g  t 2  u1 sin   t  12 g t 2
y2   u2 sin   t  12 g t 2
y2  y1   u2 sin   t  12 g t 2    u1 sin   t  12 g t 2    u2  u1  sin  t
x2  x1   u2 cos   t   u1 cos   t   u2  u1  cos  t
tan  
y2  y1  u2  u1  sin  t

 tan 
x2  x1  u2  u1  cos  t
 
The inclination angle is  hence, the inclination is independent of time.
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(b)
The inclination angle of the lining joining the stones is independent of time or position
The distance from the landing point of stone 1 at the point P to the foot of the wall is
xQ  xP  h tan 
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(c)
Show that
u2  tan   tan    2 u1 tan 
Rearranging this equation gives (A)
 2u

tan    1  1 tan 
 u2

Using the equations for constant acceleration at the time t that stone 1 hits the ground at
the point P
Stone 1 (vertical motion)
s  ut  12 a t 2
Stone 2
v  u  at
0  u1 sin  t  12 g t 2
t
2 u1 sin 
g
v2 x  u2 cos 
 2 u sin  
v2 y  u2 sin   g  1

g


v2 y   u2  2u1  sin  
tan   
v2 y
v2 x

 u2  2u1  sin  
u2 cos 
 u

tan    2 1  1 tan 
 u2

Which is the result we need to show as given by equation (A).
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QED
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(d)
  /2
 u

tan    2 1  1 tan 
 u2

 u

tan  / 2    2 1  1 tan  / 2   / 2 
 u2

 u
  tan  / 2   tan  / 2  
tan  / 2    2 1  1 

1  tan 2  / 2 
 u2


let
 u

K   2 1  1
 u2

2
z  1  2K
z  tan  / 2 
z  z3  2 K z
But z  tan  / 2 must be a real quantity
2K  1
 u

2  2 1  1  1
 u2

u
4 1 3
u2
u1  43 u2
QED
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