Question

A shell is fired from a point O at an angle of inclination of 60° with horizontal with a speed of 40 m/s and it strikes a horizontal plane through O, at a point A. Another shell is fired with the same angle of elevation but a different speed v. If it hits a drone which starts to rise vertically from A with a constant speed m/s at the same instant as the second shell is fired, find v (in m/s).

Answer 1

v = m/√3 + (1/2)√(4m²/3+6400)

Answer 2

We are given that the first shell is fired from O with speed 40 m/s at an angle 60° and lands at A. Its range is

$$R=\frac{40^2\sin 120^\circ}{g}=\frac{1600\cdot (\sqrt3/2)}{g}=\frac{800\sqrt3}{g}\,.$$

Thus, A is at a horizontal distance $R=\frac{800\sqrt3}{g}$ from O.

A second shell is fired from O with speed $v$ at the same angle 60°. Its horizontal and vertical positions (taking O as the origin) at time $t$ are

$$x(t)=v\cos60^\circ\,t=\frac{v}{2}\,t,\qquad y(t)=v\sin60^\circ\,t-\frac12gt^2=\frac{v\sqrt3}{2}\,t-\frac12gt^2\,.$$

A drone at A starts rising vertically with a constant speed (let its value be $m$ m/s) at the very instant the second shell is fired. Its vertical motion is simply

$$y_{\text{drone}}(t)=m\,t\,.$$

For the shell to hit the drone the following must hold at the collision time $t$:

1. Horizontally, the shell must cover the distance from O to A:

   $$\frac{v}{2}\,t=R=\frac{800\sqrt3}{g}\quad\Longrightarrow\quad t=\frac{1600\sqrt3}{gv}\,.$$

2. Vertically the shell and drone must be at the same height:

   $$\frac{v\sqrt3}{2}\,t-\frac12gt^2 = m\,t\,.$$

   Cancel $t\neq 0$:

   $$\frac{v\sqrt3}{2}-\frac12gt = m\,.$$

   Substitute $t=\frac{1600\sqrt3}{gv}$:

   $$\frac{v\sqrt3}{2} - \frac12g\left(\frac{1600\sqrt3}{gv}\right) = m\,.$$

   Notice that $\frac12g\cdot\frac{1600\sqrt3}{gv}=\frac{800\sqrt3}{v}$. Thus the equation becomes

   $$\frac{v\sqrt3}{2} - \frac{800\sqrt3}{v}= m\,.$$

Multiply through by 2 to eliminate fractions:

$$v\sqrt3 - \frac{1600\sqrt3}{v}=2m\,.$$

Multiply by $v$ (with $v>0$):

$$v^2\sqrt3-1600\sqrt3 = 2mv\,.$$

Rearrange to obtain a quadratic in $v$:

$$v^2\sqrt3 - 2mv - 1600\sqrt3=0\,.$$

Divide through by $\sqrt3$ (since $\sqrt3>0$):

$$v^2 - \frac{2m}{\sqrt3}\, v - 1600=0\,.$$

Using the quadratic formula,

$$v=\frac{\frac{2m}{\sqrt3}\pm\sqrt{\left(\frac{2m}{\sqrt3}\right)^2 +4\cdot1600}}{2}\,.$$

Simplify the discriminant:

$$\left(\frac{2m}{\sqrt3}\right)^2 =\frac{4m^2}{3}\quad\text{and}\quad4\cdot1600=6400\,.$$

Thus,

$$v=\frac{\frac{2m}{\sqrt3}\pm\sqrt{\frac{4m^2}{3}+6400}}{2} \,.$$

Since $v$ must be positive, we take the positive sign:

$$v=\frac{m}{\sqrt3}+\frac{1}{2}\sqrt{\frac{4m^2}{3}+6400}\,.$$