
Suppose a ball is kicked upward with an initial velocity of 10 m/s at an angle of 30° from the horizontal. How far does it travel before it hits the ground? At what angle of elevation should you kick the ball for it to reach the maximum distance? Both questions are answered by applying the equations of motion in two dimensions.
The equations of motion are the same as before (equations 6.6 and 6.7), except that we are now given the magnitude of the initial velocity v_{0} instead of the individual velocity components.
Using the projection angle, we get the initial individual velocity components of v_{x0} = v_{0} cos θ and v_{y0} = v_{0} sin θ, leading to the modified equations given below.

(6.8)  $$x={v}_{0}t\text{\u202f}\mathrm{cos}\text{\u202f}\theta $$
$${v}_{x}={v}_{0}\text{\u202f}\mathrm{cos}\text{\u202f}\theta $$
 x  =  position in xdirection (m)  v_{0}  =  initial speed (m/s)  θ  =  projection angle (degrees)  t  =  time (s)  v_{x}  =  velocity component in xdirection (m/s) 
 Projectile motion xcoordinate equations 

(6.9)  $$y={v}_{0}t\text{\u202f}\mathrm{sin}\text{\u202f}\theta {\scriptscriptstyle \frac{1}{2}}g{t}^{2}$$
$${v}_{y}={v}_{0}\text{\u202f}\mathrm{sin}\text{\u202f}\theta gt$$
 y  =  position in ydirection (m)  v_{0}  =  initial speed (m/s)  θ  =  projection angle (degrees)  t  =  time (s)  v_{y}  =  velocity component in ydirection (m/s) 
 Projectile motion ycoordinate equations 

The range of a projectile is the horizontal distance it covers before hitting the ground. That means x = v_{0}t cos θ is the equation we want to solve. The initial speed v_{0} is known, and the angle θ is also known, while the time t is not known. So we use the y equations to solve for time t. The ycomponent equation for position is a quadratic, which means there are two solutions. Factoring the equation gives the following.


The first solution is for t = 0; when the ball is first kicked at the start. The second solution is the one we want because it describes the second time the ball touches the ground. Solving the second solution for time yields t = 2v_{0} sin θ_{0} / g. Substituting this into x = v_{0}t cos θ yields the range equation:

(6.10)  $$x=\frac{2{v}_{0}^{2}\text{\u202f}\mathrm{sin}\theta \text{\u202f}\mathrm{cos}\theta}{g}$$  x  =  range (m)  v_{0}  =  initial speed (m/s)  θ  =  launch angle relative to horizontal  g  =  acceleration of gravity (9.8 m/s^{2}) 
 Range equation


What is the time for the ball to return to the ground when it is thrown directly upward at 18 m/s?

Using t = x/v_{0} cos θ_{0} = 2v_{0} sin θ_{0}/g
and realizing that “upward” means that θ = 90°, we substitute values to get
$$t=\frac{2{v}_{0}\mathrm{sin}\theta}{g}=\frac{2(18\text{m/s})\mathrm{\u202fsin}\mathrm{\u202f90}\xb0}{9.8{\text{m/s}}^{2}}=3.7\text{s}$$
