Ncert Exercises & Solutions

4.30. A gun can fire shells with maximum speed $v_{0}$ and the maximum horizontal range that can be achieved is $R = \frac{v_{0}^{2}}{g}$ . If a target farther away by distance ∆x has to be hit with the same gun, show that it could be achieved by raising the gun to a height at least $h = Δ x [1 + \frac{Δ x}{R}]$

This problem can be approached in two different ways

(i) Refer to the diagram, target T is at a horizontal distance $x = R + Δ x$ and between the point of projection y = - h.

(ii) From point P in the diagram projection at speed $v_{0}$ at an angle $θ$ below horizontal with height h and horizontal range ( $∆$ x)

Hint: Time is taken to travel horizontal and vertical distances will be the same.

Step 1: Find the maximum range.
$R = \frac{v_{0}^{2}}{g}, for θ = 45^{\circ} . . . (i)$

Step 2: Apply equation in the horizontal direction

The horizontal component of initial velocity = $v_{0} \cos θ$

$\begin{array}{l} (R + Δx) = v_{0} cosθ \times t \\ \Rightarrow t = \frac{(R + Δx)}{v_{0} cosθ} \end{array}$ ...(ii)

Step 3: Put the value of t in from equation (ii) for vertical motion and calculate h

Let the gun be raised through a height h from the ground so that it can hit the target. Let vertically downward direction is taken as positive

The vertical component of initial velocity $= - v_{0} \sin θ$
Taking motion in the vertical direction, $h = (- v_{0} \sin θ) t + \frac{1}{2} g t^{2}$ ...(iii)

Substituting the value of t in Eq. (iii), we get
$\begin{array}{l} h = (- v_{0} sinθ) \times (\frac{R + Δx}{v_{0} cosθ}) + \frac{1}{2} g (\frac{R + Δx}{v_{0} cosθ})^{2} \\ h = - (R + Δx) tanθ + \frac{1}{2} g \frac{(R + Δx)^{2}}{v_{0}^{2} \cos^{2} θ} \end{array}$
As the angle of projection is $θ = 45 °$ , therefore

$\begin{array}{l} h = - (R + Δx) + \tan 45^{\circ} + \frac{1}{2} g \frac{(R + Δx)^{2}}{v_{0}^{2} \cos^{2} 45^{\circ}} \\ h = - (R + Δx) \times 1 + \frac{1}{2} g \frac{(R + Δx)^{2}}{v_{0}^{2} (1 / 2)} \end{array}$
$(∵ \tan 45^{\circ} = 1 and \cos 45^{\circ} = \frac{1}{\sqrt{2}})$
$\begin{array}{l} h = - (R + Δx) + \frac{(R + Δx)^{2}}{R} [using EQ . (i) R = v_{0}^{2} / g] \\ = - (R + Δx) + \frac{1}{R} (R^{2} + {Δx}^{2} + 2 RΔx) \\ = - R - Δx + (R + \frac{{Δx}^{2}}{R} + 2 Δx) \\ = Δx + \frac{{Δx}^{2}}{R} \\ h = Δx (1 + \frac{Δx}{R}) \end{array}$