3 MOTION IN 3 DIMENSIONS 3.12 Relative velocity
×qv + v +2 g z (^0 0)
Worked example 3. 2 : Cannon shot
Question: A cannon placed on a 50 m high cliff fires a cannonball over the edge
of the cliff at v = 200 m/s making an angle of θ = 30 ◦ to the horizontal. How
long is the cannonball in the air? Neglect air resistance.
Answer: In order to answer this question we need only consider the cannon-
ball’s vertical motion. At t = 0 (i.e., the time of firing) the cannonball’s height
off the ground is z 0 = 50 m and its velocity component in the vertical direction is
v 0 = v sin θ = 200 sin 30 ◦ = 100 m/s. Moreover, the cannonball is accelerating
vertically downwards at g = 9.91 m/s^2. The equation of vertical motion of the
cannonball is written
z = z 0 + v 0 t −
1
g t^2 ,
2
where z is the cannonball’s height off the ground at time t. The time of flight of
the cannonball corresponds to the time t at which z = 0. In other words, the time
of flight is the solution of the quadratic equation
Hence,
0 = z 0 + v 0
t −
1
g t^2.
2
2
t = 0
g
Here, we have neglected the unphysical
negative root of our quadratic equation.
= 20.88 s.