4.3 Projectile motion  (Page 3/13)

Because $y_0$ and $v_y$ are both zero, the equation simplifies to

Solving for y gives

Now we must find $v_{0y},$ the component of the initial velocity in the y direction. It is given by $v_{0y}=v_0\text{sin}{\theta }_0,$ where $v_0$ is the initial velocity of 70.0 m/s and ${\theta }_0=75\text{°}$ is the initial angle. Thus,

and y is

Thus, we have

Note that because up is positive, the initial vertical velocity is positive, as is the maximum height, but the acceleration resulting from gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6-m/s initial vertical component of velocity reaches a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, so the initial velocity would have to be somewhat larger than that given to reach the same height.

(b) As in many physics problems, there is more than one way to solve for the time the projectile reaches its highest point. In this case, the easiest method is to use $v_y=v_{0y}-gt.$ Because $v_y=0$ at the apex, this equation reduces to simply

or

This time is also reasonable for large fireworks. If you are able to see the launch of fireworks, notice that several seconds pass before the shell explodes. Another way of finding the time is by using $y\text{=}{y}_0+\frac{1}{2}(v_{0y}+v_y)t.$ This is left for you as an exercise to complete.

(c) Because air resistance is negligible, $a_x=0$ and the horizontal velocity is constant, as discussed earlier. The horizontal displacement is the horizontal velocity multiplied by time as given by $x=x_0+v_{x}t,$ where $x_0$ is equal to zero. Thus,

where $v_x$ is the x -component of the velocity, which is given by

Time t for both motions is the same, so x is

Horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. When the shell explodes, air resistance has a major effect, and many fragments land directly below.

(d) The horizontal and vertical components of the displacement were just calculated, so all that is needed here is to find the magnitude and direction of the displacement at the highest point:

Note that the angle for the displacement vector is less than the initial angle of launch. To see why this is, review [link] , which shows the curvature of the trajectory toward the ground level.

When solving [link] (a), the expression we found for y is valid for any projectile motion when air resistance is negligible. Call the maximum height y = h . Then,

This equation defines the maximum height of a projectile above its launch position and it depends only on the vertical component of the initial velocity.