Motion in Two or Three Dimensions - Projectile Motion

One of the simplest and most elegant examples of this an object under the influence of gravity. Suppose a cannonball is shot upward with a speed of 10.0 m/s at an angle of 30.0 degrees to the horizontal. The cannonball is at a height of 15 meters. We are looking for the x distance traveled by the cannon. As said before in the vector section, orthogonal vectors do not affect each other. Using this, it make sense to change the vector to rectangular coordinates and then use the individual components.

This is the initial x and y velocity. Gravity affects the ball with a constant acceleration in the y direction only. The acceleration due to gravity near the earth's surface is 9.8 m/s^2 downward or
-9.8 m/s. Since there is a constant acceleration, it makes sense to use this equation.

Plugging in:

Solving for this using the quadratic formula, two values of time are gotten. They are 2.3 and -1.3. As time cannot be negative the second value is not used. Thus, the time it takes for the ball to hit the ground is 2.3 seconds. As no other acceleration affects the object, the x velocity is constant. Thus multiplying the time taken by the object to hit the ground is 8.66*2.3 or 20.2 meters.