Irodov 1.32: One Range, Two Flight Times – Amit & Deepali Sinha Foundation

Problem Irodov 1.32

A cannon and a target sit on the same horizontal level, a distance $R=5.10$ km apart. A shell is fired from the cannon with initial speed $v_0=240$ m/s, air resistance negligible; the launch angle is not fixed by the problem, only the requirement that the shell land on the target. Find how long the shell is in flight before it reaches the target.

The setup: the cannon at O fires with muzzle speed v_0 at a launch angle θ the problem leaves open, and the shell must land on the target at range R on the same level. The arc is drawn at an arbitrary θ. — The setup: the cannon at $O$ fires with muzzle speed $v_0$ at a launch angle $\theta$ the problem leaves open, and the shell must land on the target at range $R$ on the same level. The arc is drawn at an arbitrary $\theta$ .

Given

$R = 5.10\text{ km}$
$v_0 = 240\text{ m/s}$

Find

the flight time

t

to the target

Solution

Step 1Range as a function of the launch angle

With initial speed $v_0$ at angle $\theta$ above the horizontal and constant downward acceleration $g$ , integrating once from the launch conditions gives the coordinates at any time.

x(t) = v_0\cos\theta\,t, \qquad y(t) = v_0\sin\theta\,t-\frac{gt^2}{2}

The shell returns to the launch height when $y$ vanishes a second time; solving for the nonzero root gives the flight time, and substituting it into $x(t)$ gives the range, using the double angle identity on the way.

\begin{aligned} 0 &= v_0\sin\theta\,t_{\mathrm{flight}}-\frac{gt_{\mathrm{flight}}^2}{2} \\ t_{\mathrm{flight}} &= \frac{2v_0\sin\theta}{g} \end{aligned} \tag{1}

R = x(t_{\mathrm{flight}}) = v_0\cos\theta\cdot\frac{2v_0\sin\theta}{g} = \frac{v_0^2(2\sin\theta\cos\theta)}{g}

R = \dfrac{v_0^2\sin2\theta}{g}

(2)

Step 2The two launch angles that reach the target

What this step establishes: two trajectories reach the same target, the flatter one launched at θ_1 and the lofted one at θ_2=90°-θ_1. Both land at range R because 2θ_1=2θ_2, and the flight times differ. — What this step establishes: two trajectories reach the same target, the flatter one launched at $\theta_1$ and the lofted one at $\theta_2=90^\circ-\theta_1$ . Both land at range $R$ because $\sin2\theta_1=\sin2\theta_2$ , and the flight times differ.

Solving (2) for $\theta$ gives two roots in $(0^\circ,90^\circ)$ , related as complements.

\sin2\theta = \frac{gR}{v_0^2}

2\theta = \arcsin\!\left(\frac{gR}{v_0^2}\right) \quad\text{or}\quad 2\theta = 180^\circ-\arcsin\!\left(\frac{gR}{v_0^2}\right)

\theta_0 = \frac12\arcsin\!\left(\dfrac{gR}{v_0^2}\right)

(3)

so the two launch angles are $\theta_1=\theta_0$ (flat) and $\theta_2=90^\circ-\theta_0$ (lofted).

Step 3Flight time for each angle

The flight time (1), evaluated at $\theta_0$ and at $90^\circ-\theta_0$ with $\sin(90^\circ-\theta_0)=\cos\theta_0$ , gives the two times directly.

t_1 = \frac{2v_0}{g}\sin\theta_0, \qquad t_2 = \frac{2v_0}{g}\sin(90^\circ-\theta_0) = \frac{2v_0}{g}\cos\theta_0 \tag{4}

Substituting $R=5100$ m, $v_0=240$ m/s and $g=9.8$ m/s $^2$ into (3) evaluates the split angle.

\theta_0 = \frac12\arcsin\!\left(\frac{9.8\times5100}{240^2}\right) = \frac12\arcsin(0.868) = 30.1^\circ

Substituting $\theta_0=30.1^\circ$ into (4) gives the flight times in seconds.

\begin{align*}t_1 &= \frac{2(240)}{9.8}\sin30.1^\circ = 24.6\ \text{s} \\ t_2 &= \frac{2(240)}{9.8}\cos30.1^\circ = 42.4\ \text{s}\end{align*}

t_1 = \dfrac{2v_0}{g}\sin\theta_0 = 24.6\ \text{s} = 0.41\ \text{min}

(5)

t_2 = \dfrac{2v_0}{g}\cos\theta_0 = 42.4\ \text{s} = 0.71\ \text{min}

(6)

Check

The product $t_1t_2=(2v_0/g)^2\sin\theta_0\cos\theta_0=(2v_0^2/g^2)\sin2\theta_0=(2v_0/g)(R/v_0)=2R/g=2(5100)/9.8=1041\ \text{s}^2$ , and indeed $24.6\times42.4=1043\ \text{s}^2$ , a tight match.

Remark

As $R$ approaches the maximum range $v_0^2/g$ , $\theta_0$ approaches $45^\circ$ and $t_1,t_2$ merge into a single flight time: the maximum-range trajectory is exactly the launch angle where the two roots of Step 2 coincide.