
How can we simplify these equations of motion for projectiles? We use the fact that gravity accelerates down; therefore, a_{y} = –g and a_{x} = 0! The equations of motion simplify a great deal because all the terms that include a_{x} become zero. Also, to make the math simpler, we choose the initial position to be zero, so x_{0} = 0 m and y_{0} = 0 m.

Let’s first examine the motion in the xdirection. Since there is no acceleration in x, the component of velocity in the xdirection is constant. The projectile moves with constant velocity along x! The x position changes linearly, as shown in equation (6.6).

(6.6)  $$x={v}_{x0}t$$
$${v}_{x}={v}_{x0}$$
 x  =  position in xdirection (m)  v_{x}_{0}  =  initial velocity in xdirection (m/s)  v_{x}  =  velocity component in xdirection (m/s)  t  =  time (s) 
 Projectile motion xcomponent 

Now look at the equations of motion in the ydirection. The acceleration in y has a constant value of −g. As we learned in Chapter 4, constant (nonzero) acceleration results in velocity that changes linearly, as shown in equation (6.7). Position in the ydirection is a function of t and t^{2}, as also shown in equation (6.7).

(6.7)  $$y={v}_{y0}t{\scriptscriptstyle \frac{1}{2}}g{t}^{2}$$
$${v}_{y}={v}_{y0}gt$$
 y  =  position in ydirection (m)  v_{y}_{0}  =  initial velocity in ydirection (m/s)  v_{y}  =  velocity component in ydirection (m/s)  t  =  time (s)  g  =  acceleration due to gravity = 9.8 m/s^{2} 
 Projectile motion ycomponent 

Equations (6.6) and (6.7) are the equations of projectile motion when gravity is acting in the negative ydirection. The initial velocity given to the projectile—soccer ball, cannonball, etc.—has components v_{x0} in the xdirection and v_{y0} in the ydirection.


Consider a baseball player hitting a ball with an upward velocity of 30 m/s at an angle of 60°above horizontal. The ball’s initial velocity has components of 15 and 26 m/s in the x and ydirections, respectively. Applying equations (6.6) and (6.7) predicts the trajectory illustrated above—with the magnitude of the velocity calculated from v^{2} = v_{x}^{2} + v_{y}^{2}.

In the illustration of the fly ball above, the position of the ball is depicted at fixed time intervals (every 0.5 s) to create a particle model of its motion. The ball positions are closest to each other at the top of the trajectory, which means that the speed is slowest there. The ball positions are furthest apart near the ground where the speed is fastest.

When the baseball is first hit, its velocity has a position angle of 60° with the horizontal. After 2 s, what is the position angle of its velocity?

From the illustration above, after 2 s the magnitude of the velocity is v = 16.3 m/s. We already know that the xcomponent of the velocity is constant at v_{x} = 15 m/s, because there is no acceleration acting in the xdirection. Since we know the total velocity (the hypotenuse of the vector triangle) and the velocity in the xdirection, then we can get the position angle using the inverse cosine function: $$\theta ={\mathrm{cos}}^{1}\left(\frac{{v}_{x}}{v}\right)={\mathrm{cos}}^{1}\left(\frac{15\text{m/s}}{16.3\text{m/s}}\right)=23.0\xb0$$
