Calculus Of One Real Variable –
By Pheng Kim Ving 
6.1.6 
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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.
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Fig. 2.1
Trajectory Of Projectile Motion.
The arrow head indicates the direction
of motion.

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:
Fig. 2.2

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 = Ax^{2} + Bx where A and
B are
certain constants, the trajectory is a parabola.
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3. Finding Maximum Height, Flight Time, And Range 
The maximum height is the maximum vertical position y_{max} 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 t_{max} 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 x_{max} attained by
the object; it's the horizontal distance between the firing and impact points.
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 35^{o}
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/sec^{2}.
Solution
Let s_{0} be the muzzle speed and t_{max} the flight time of the shell. The range is:
2 km = 2000 m = x_{max} = (s_{0} cos 35^{o})t_{max},
so that:
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