## A Lesson in Ballistics

In the previous lesson, we saw how one can calculate the vertical position of an object as a function of its initial velocity and the accelerationofgravity. In this lesson, we will extend that concept to allow us to calulate the horizontal position of the object as well. To demonstrate this we will use a common application of this principle: ballistics.

For our example we will use the cannon shown on the left. Our cannon fires a cannonball. This cannonball
leaves the barrel of the cannon at some intial velocity V_{0}. Recall that velocity is the
vector that includes both the speed and the direction of an object. So initially, the cannonball has a
velocity that includes both a horizontal component, V_{x} and a vertical component,
V_{y}.

From trigonometry, it is easy to see that we can calulate the components from the velocity:

V

_{x}= V_{0}* cos(θ)V

_{y}= V_{0}* sin(θ)

where θ is the angle of the initial trajectory of the cannonball.

The important point to grasp in this lesson is, as in the previous lesson, the forces acting on the object
to not change with time (*as before, we ignore the effects of friction from the atmosphere*).

Therefore, the position of the cannonball at any time after time t_{0} can be calculated as:

x

_{t}= V_{x}* ty

_{t}= V_{y}* t - 1/2 * g * t^{2}

where t is the time in seconds from the firing the cannon and g is the acceleration of gravity
(9.802m/s^{2}).

Questions:

- If the initial velocity is 15m/s and the barrel angle is 50o above the horizon, what is the maximum height the cannonball will reach? (mouse over here for a hint).
- In the above example, how far will the ball travel before it hits the ground (assume the “ground”is the faint brown line, which is the elevation the cannon ball starts at time zero)? (mouse over here for a hint)