PASCO ME-6831 Ballistic Pendulum_Projectile Launcher - Manual (Page 59)

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M o d e l N o . M E - 6 8 3 0 T e a c h e r ’ s G u i d e
0 1 2 - 0 5 3 7 5 C 55
• substitute
• Substitute
• and simplify:
• When x = x
max
, the value of this derivative is negative.
• Therefore, the ball has already reached the peak and is on its way down.
2. Solve the equation for maximum angle to determine x:
• Substitute this value into the equation for y to determine the maximum height.
Exp. 8: Projectile Motion—Approximate Method
Procedure
1. The exact mass used is not critical. Pick a value that gives a fairly large swing for best results.
2. With the steel ball and extra masses on the pendulum, the balance point will be somewhere on the ball catcher
itself. This makes it difficult to use string, but it is relatively easy to find the center of mass by balancing the
pendulum on a straightedge.
3. The angle reached by the pendulum should not vary by more than one degree between successive trials.

max
v
0
2
gx
max
--------------
 
 
 
1 –
tan =
dy
dx
------
v
0
2
gx
max
--------------
gx
v
0
2
v
0
2
gx
max
--------------
 
 
 
1 –
tan
 
 
 
2
cos
-------------------------------------------------------- – =
a
b
---  
  1 –
tan  
 
cos
b
a
2
b
2
+
--------------------- =
dy
dx
------
v
0
2
gx
max
--------------
gx
v
0
2
gx
max
v
0
4
g
2
x
max
2
+
---------------------------------
 
 
  2
----------------------------------------------- –
v
0
2
gx
max
--------------
x v
0
4
g
2
x
2
max +
 
v
0
2
gx
max
2
--------------------------------------- – = =
dy
dx
------
v
0
gx
max
--------------
v
0
x
gx
2
max
----------------- –
gx
v
0
2
------ – =
dy
dx
------
x
max
gx
max
v
0
2
-------------- – =

max
tan
v
0
2
gx
------ x

v
0
2
g
----- = =
y y
0
v
0
2
g
-----
g
v
0
2
g
-----
 
 
 
2
v
0
2
---------------- – + y
0
v
0
2
g
-----
v
0
2
g
----- – + = =
y y
0
=