Range of a ProjectileNo Air Resistance If you worked in Exercises 3.12, you saw that

Chapter 4, Problem 21

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Range of a Projectile—No Air Resistance If you worked Problem 23 in Exercises 3.12, you saw that when air resistance and all other forces except its weight w  mg are ignored, the path of motion of a ballistic projectile, such as a cannon shell, is described by the system of linear differential equations

\(m \frac{d^{2} x}{d t^{2}}=0\)

\(m \frac{d^{2} y}{d t^{2}}=-m g\)

(a) If the projectile is launched from level ground with an initial velocity \(v_{0}\) assumed to be tangent to its path of motion or trajectory, then the initial conditions accompanying the system are x(0)= 0, x’(0)= \(v_{0} \cos \theta\), y(0)= 0, y’(0)= \(v_{0} \sin \theta\) where\(v_{0}=\left||\mathbf{v}_{0}\right||\) the initial speed is constant and \(\theta\) is the constant angle of elevation of the cannon. See Figure 3.R.5 in The Paris Guns problem on page 206. Use the Laplace transform to solve system (8).

(b) The solutions x(t) and y(t) of the system in part (a) are parametric equations of the trajectory of the projectile. By using x(t) to eliminate the parameter t in y(t), show that the trajectory is parabolic.

(c) Use the results of part (b) to show that the horizontal range R of the projectile is given by

\(R=\frac{v_{0}^{2}}{g} \sin 2 \theta\)

From (9) we not only see that R is a maximum when \(\theta\)= \(\pi\)/4 but that a projectile launched at distinct complementary angles \(\theta\) and \(\pi\)/2-\(\theta\) has the same sub maximum range. See FIGURE 4.6.8. Use a trigonometric identity to prove this last result.

(d) Show that the maximum height H of the projectile is given by

\(H=\frac{v_{0}^{2}}{2 g} \sin ^{2} \theta\)

(e) Suppose g= 32 \(\mathrm{ft} / \mathrm{s}^{2}\) , \(\theta\)= \(38^{\circ}\), and \(v_{0}\)= 300 ft/s. Use (9) and (10) to find the horizontal range and maximum height of the projectile. Repeat with \(\theta=52^{\circ}\) and \(v_{0}\)= 300 ft/s.

(f ) Because formulas (9) and (10) are not valid in all cases (see Problem 22), it is advantageous to you to remember that the range and maximum height of a ballistic projectile can be obtained by working directly with x(t) and y(t), that is, by solving y(t)= 0 and y’(t)= 0. The first equation gives the time when the projectile hits the ground and the second gives the time when y(t) is a maximum. Find these times and verify the range and maximum height obtained in part (e) for the trajectory with \(\theta\)= \(38^{\circ}\) and \(v_{0}\)- 300 ft/s. Repeat with \(\theta=52^{\circ}\).

(g) With g= 32 \(\mathrm{ft} / \mathrm{s}^{2}\), \(\theta\)= \(38^{\circ}\)  and \(v_{0}\)= 300 ft/s use a graphing utility or CAS to plot the trajectory of the projectile defined by the parametric equations x(t) and y(t) in part (a). Repeat with \(\theta=52^{\circ}\). Using different colors superimpose both curves on the same coordinate system.