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Calculus 1 Problems & Solutions  –  Chapter 4  –  Section 4.1.1.6

4.1.1.6 The Projectile Motion

 

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Review

 

1. The Projectile Motion

 

In Section 3.8 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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2. Equations Of The Trajectory

 

 

Fig. 2.1   Trajectory Of Projectile Motion.   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 3.8 Part 3 we get x'' = 0. The horizontal velocity is x'(t).

position is x(0) = 0. Hence, we obtain this initial-value problem (see Section 3.9 Part 4):

 

 

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 3.8 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 Aand 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.

 

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.

 

 

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Problems & Solutions

 

 

 

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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/sec².

 

Solution

 

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

 

2 km = 2000 m = x_max = (s₀ cos 35^o)t_max,

 

so that:

 

 

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