V0
+y

+x
h
x
V0
In the x-direction we only have one
equation:
X = (V0 cos ) t
+y
+x
As you can see we need to solve for the
time, t. We can use the motion in the ydirection to accomplish this.
H = (V0 sin ) t + ½ g t2
h
x
If we take H to the other side:
½ gt2 + (V0 sin ) t – H = 0
This is a quadratic equation so, we have to use the quadratic formula to solve for t:
1
 V0 sin   (V0 sin  )  4( g )(  H )
2
t
1
2( g )
2
2
V0
Now that we have the time t we can find the range
By using: X = (V0 cos ) t
+y
+x
h
x
For this special case how will we find the muzzle
velocity of the cannon?
Remember in the x-direction:
V0
X = (V0 cos ) t = V0 t so V0 = X/t
In the y-direction:
H = (V0sin )t + ½ gt2 = ½ gt2
If we solve for t we get:
t
2H
g
g
 V0  X
2H
h
x
Today you are going to first measure the muzzle velocity of the cannon.
To do this you put the cannon at  = 0 and measure the height (H) and fire
the cannon a few times. Each time measure the range (X) then plug the
average value for X into:
g
V0  X
2H
This is the muzzle velocity of the cannon. Next you will use
your calculations to predict where the ball will land at
different angles. (See the table on page 21)
Finally I will give you a bull’s eye and I will write the angle
down on the sheet. You will predict where the ball will land
and try to see if you can hit the bull’s eye. (Points will be
rewarded accordingly).