$$\overrightarrow r=\overrightarrow{v_0}t+\frac{1}{2}\overrightarrow{g}t^2$$
where \(\overrightarrow r\) is the position of the ball at time \(t\), \(\overrightarrow{v_0}\) is the initial velocity of the ball, \(\overrightarrow{g}\) is the acceleration due to gravity, and \(t\) is the time.
This equation is valid for any object moving in two dimensions under the influence of gravity, regardless of the direction in which it is thrown. The only restriction is that the object must be moving in a plane parallel to the ground.
To see how the equation of projectile motion applies to a ball thrown in an arbitrary direction, let's consider the following example. Suppose a ball is thrown with an initial velocity of 10 m/s at an angle of 30 degrees above the horizontal. The equation of projectile motion for this ball is:
$$\overrightarrow r=(10\cos30^\circ)\hat{i}+(10\sin30^\circ)t\hat{j}-\frac{1}{2}gt^2\hat{j}$$
where \(\hat{i}\) and \(\hat{j}\) are the unit vectors in the horizontal and vertical directions, respectively.
This equation can be used to calculate the position of the ball at any time \(t\). For example, at time \(t = 1\text{ s}\), the position of the ball is:
$$\overrightarrow r=(10\cos30^\circ)\hat{i}+(10\sin30^\circ)(1\text{ s})\hat{j}-\frac{1}{2}(9.8\text{ m/s}^2)(1\text{ s})^2\hat{j}$$
$$=(8.66\text{ m})\hat{i}+(5\text{ m})\hat{j}-(4.9\text{ m})\hat{j}$$
$$=(8.66\text{ m})\hat{i}+(0.1\text{ m})\hat{j}$$
Thus, the ball is located 8.66 m from the starting point in the horizontal direction and 0.1 m from the starting point in the vertical direction.
The equation of projectile motion can be used to solve a variety of problems involving the motion of objects under the influence of gravity. For example, it can be used to calculate the range of a projectile, the maximum height of a projectile, and the time of flight of a projectile.
