Calculus Of One Real Variable – By Pheng Kim Ving Chapter 6: The Trigonometric Functions And Their Inverses – Section 6.1.6: The Projectile Motion 6.1.6 The Projectile Motion

 1. The Projectile Motion

In Section 5.8 we studied the motion of an object moving along a path that is a line. In this section we'll study the motion Fig. 2.1. The path along which the object moves is called the trajectory of the object.

 2. Equations Of The Trajectory  # The arrow head indicates the direction of motion.

### Parametric Equations

Clearly x and y are both functions of time t: x = x(t) and y = y(t). First, let's find x(t), ie express x as a function of t.
During the object's flight, it experiences no horizontal force acting on it, assuming that air resistance is negligible. So, by its horizontal acceleration, we must have a = 0. Thus, by Section 5.8 Part 3 we get x'' = 0. The horizontal velocity is x'(t). position is x(0) = 0. Hence we obtain the following system of equations:  # Next let's find y(t). Since we assume that air resistance is negligible, there's only one vertical force acting on the object
during its flight; that force is gravity. It follows that, by Section 5.8 Eq. [6.1], we get:  That's the Cartesian equation of the trajectory of the object. Since it's of the form y = Ax2 + Bx where A and B are
certain constants, the trajectory is a parabola.

 3. Finding Maximum Height, Flight Time, And Range

The maximum height is the maximum vertical position ymax attained by the object. Refer to Fig. 2.1. The flight time is
the time the flight of the object takes to complete; since the initial time is 0, it's the time tmax when the object strikes the
ground. The point where the object strikes the ground is also called the impact point. The range is the maximum
horizontal  position xmax attained by the object; it's the horizontal distance between the firing and impact points.

### Example 3.1  EOS

Remarks 3.1

a.  We utilize the parametric equations [2.1] and [2.2] to find the maximum height, flight time, and range. Problems & Solutions   2.  A shell is fired at an angle 35o above the horizontal and strikes the ground at a point 2 km horizontally away from its
firing point. Find the muzzle speed of the shell. The acceleration due to gravity is g = 9.8 m/sec2.

Solution

Let s0 be the muzzle speed and tmax the flight time of the shell. The range is:

2 km = 2000 m = xmax = (s0 cos 35o)tmax,

so that: 