Irodov 1.33: Timing the Second Shot to Collide – Amit & Deepali Sinha Foundation

Problem Irodov 1.33

A cannon fires two shells one after the other from the same point, with the same azimuth (so both flights stay in one vertical plane) and the same launch speed $v_0=250$ m/s. The first shell is launched at $\theta_1=60^\circ$ above the horizontal, the second at $\theta_2=45^\circ$ . Air resistance is negligible. Find the time interval $\Delta t$ between the two firings for which the shells collide in flight.

The setup: both shells leave the same point with the same speed v_0, the first at θ_1=60° and the second at θ_2=45°, so their paths lie in one vertical plane. — The setup: both shells leave the same point with the same speed $v_0$ , the first at $\theta_1=60^\circ$ and the second at $\theta_2=45^\circ$ , so their paths lie in one vertical plane.

Given

$v_0 = 250\text{ m/s}$
$\theta_1=60^\circ$ , $\theta_2=45^\circ$

Find

the time interval

\Delta t

between the two firings

Solution

Step 1Positions of both shells on one clock

What this step demands: the two paths meet at the point C, and a collision needs both shells to be at C at the same instant of the single clock. — What this step demands: the two paths meet at the point $C$ , and a collision needs both shells to be at $C$ at the same instant of the single clock.

Let $t_1$ be the first shell’s elapsed flight time at the moment of collision and $t_2$ the second shell’s; since the second is launched a time $\Delta t$ after the first and both are airborne at the collision instant,

t_1 = t_2+\Delta t.

Each shell follows the same projectile law from the common launch point, so at the collision instant their coordinates must agree.

\begin{aligned} v_0\cos\theta_1\,t_1 &= v_0\cos\theta_2\,t_2 \\ v_0\sin\theta_1\,t_1-\frac{gt_1^2}{2} &= v_0\sin\theta_2\,t_2-\frac{gt_2^2}{2} \end{aligned} \tag{1}

Step 2Solve the horizontal equation for

t_1

and

t_2

in terms of

\Delta t

The horizontal equation of (1), with $t_1=t_2+\Delta t$ substituted, is linear in $t_2$ and gives both times as multiples of $\Delta t$ .

\begin{align*}\cos\theta_1(t_2+\Delta t) &= \cos\theta_2\,t_2 \\ t_2(\cos\theta_2-\cos\theta_1) &= \Delta t\cos\theta_1\end{align*}

t_2 = \frac{\Delta t\cos\theta_1}{\cos\theta_2-\cos\theta_1}, \qquad t_1 = t_2+\Delta t = \frac{\Delta t\cos\theta_2}{\cos\theta_2-\cos\theta_1} \tag{2}

Step 3Solve the vertical equation for

\Delta t

First the vertical equation of (1) is rearranged so that the two quadratic terms combine into a difference of squares, which factors and lets $t_1-t_2=\Delta t$ be taken out.

\begin{align*}\sin\theta_1\,t_1-\sin\theta_2\,t_2 &= \frac{g}{2v_0}(t_1^2-t_2^2) \\ &= \frac{g}{2v_0}(t_1-t_2)(t_1+t_2) \\ &= \frac{g}{2v_0}\Delta t(t_1+t_2)\end{align*}

Now both sides are turned into multiples of $\Delta t$ by putting the times (2) in. On the left, the two sine-cosine products collect into the sine of the angle difference.

\begin{align*}\sin\theta_1\,t_1-\sin\theta_2\,t_2 &= \sin\theta_1\cdot\frac{\Delta t\cos\theta_2}{\cos\theta_2-\cos\theta_1}-\sin\theta_2\cdot\frac{\Delta t\cos\theta_1}{\cos\theta_2-\cos\theta_1} \\ &= \frac{\Delta t(\sin\theta_1\cos\theta_2-\cos\theta_1\sin\theta_2)}{\cos\theta_2-\cos\theta_1} \\ &= \frac{\Delta t\sin(\theta_1-\theta_2)}{\cos\theta_2-\cos\theta_1}\end{align*}

On the right, the same substitution adds the two times.

\begin{align*}t_1+t_2 &= \frac{\Delta t\cos\theta_2}{\cos\theta_2-\cos\theta_1}+\frac{\Delta t\cos\theta_1}{\cos\theta_2-\cos\theta_1} \\ &= \frac{\Delta t(\cos\theta_1+\cos\theta_2)}{\cos\theta_2-\cos\theta_1}\end{align*}

Equating the two sides and canceling the common factor $\Delta t/(\cos\theta_2-\cos\theta_1)$ leaves one equation in $\Delta t$ alone.

\sin(\theta_1-\theta_2) = \frac{g\Delta t}{2v_0}(\cos\theta_1+\cos\theta_2)

\Delta t = \dfrac{2v_0}{g}\cdot\dfrac{\sin(\theta_1-\theta_2)}{\cos\theta_1+\cos\theta_2}

(3)

Step 4Evaluate

Substituting the given speed and angles into (3) gives the numerical delay.

\Delta t = \frac{2(250)}{9.8}\cdot\frac{\sin15^\circ}{\cos60^\circ+\cos45^\circ} = 51.02\times\frac{0.2588}{0.5+0.7071}

\Delta t = \dfrac{2v_0}{g}\cdot\dfrac{\sin(\theta_1-\theta_2)}{\cos\theta_1+\cos\theta_2} = 11\ \text{s}

(4)

Check

$\theta_1>\theta_2$ makes $\Delta t>0$ , as it must: the steeper shell has to be launched first so the flatter, later shell has a shorter flight to reach the same crossing point. Reversing the labeling ( $\theta_1=45^\circ$ fired first) would flip the sign, correctly ruling out that firing order.

Remark

The two shells collide at the one point $C$ where their trajectories, drawn on the same axes from the same launch point, physically cross; $\Delta t$ is not found by chasing that crossing point directly, but by requiring that the elapsed flight times differ by exactly the firing delay while both land on it, which is what turns a geometric intersection into a timing condition.