Question

A wall OP is inclined to the horizontal ground at an angle $\alpha$. Two particles are projected from points A and B on the ground with same speed (u) in directions making an angle $\theta$ to the horizontal (see figure). Distance between points A and B is $x_0$ = 24 m. Both particles hit the wall elastically and fall back on the ground. Time of flight (time required to hit the wall and then fall back on to the ground) for particles projected from A and B are 4 s and 2 s respectively. Both the particles strike the wall perpendicularly and at the same location. [In elastic collision, the velocity component of the particle that is perpendicular to the wall gets reversed without change in magnitude [g = 10 m/s²]

• A. Maximum height attained by the particle projected from A is 11.25 m
• B. The inclination of the wall to the horizontal ($\alpha$) is related as $tan \alpha = (\frac{8}{5})$
• C. Maximum height attained by the particle projected from A is 22.5 m
• D. The inclination of the wall to the horizontal ($\alpha$) is related as $tan \alpha = (\frac{5}{8})$

Answer 1

Answer: (A), (B)

A and B

Answer 2

Let $t_A'$ and $t_B'$ be the times taken by particles A and B, respectively, to hit the wall.
Let $(x_c, y_c)$ be the common collision point.

1. Horizontal Motion to Wall:

   $x_c = x_A + (u \cos \theta) t_A'$
   $x_c = x_B - (u \cos \theta) t_B'$
   Equating $x_c$: $(u \cos \theta) (t_A' + t_B') = x_B - x_A = 24$ m. (Eq 1)

2. Vertical Motion to Wall:

   $y_c = (u \sin \theta) t_A' - \frac{1}{2} g (t_A')^2$
   $y_c = (u \sin \theta) t_B' - \frac{1}{2} g (t_B')^2$
   Equating $y_c$: $u \sin \theta (t_A' - t_B') = \frac{1}{2} g ((t_A')^2 - (t_B')^2) \implies u \sin \theta = \frac{1}{2} g (t_A' + t_B')$. (Eq 2)

3. Perpendicular Collision Condition:

   The velocity vector $\vec{v}=(v_x, v_y)$ must be perpendicular to the wall. The wall has slope $\tan \alpha$. So, $v_x \cos \alpha + v_y \sin \alpha = 0$.
   For particle A: $(u \cos \theta) \cos \alpha + (u \sin \theta - g t_A') \sin \alpha = 0 \implies u \cos(\theta - \alpha) = g t_A' \sin \alpha$. (Eq 3A)
   For particle B: $(-u \cos \theta) \cos \alpha + (u \sin \theta - g t_B') \sin \alpha = 0 \implies -u \cos(\theta + \alpha) = g t_B' \sin \alpha$. (Eq 3B)

4. Velocity after Collision:

   The component of velocity perpendicular to the wall reverses. The component parallel to the wall is zero (due to perpendicular impact) and remains zero.
   The velocity component perpendicular to the wall before collision for A is $v_{perp,A} = \vec{v}_A \cdot \vec{n} = (u \cos \theta)\sin \alpha - (u \sin \theta - g t_A')\cos \alpha = \frac{u \cos \theta}{\sin \alpha}$.
   After collision, $v'_{perp,A} = -\frac{u \cos \theta}{\sin \alpha}$.
   Similarly for B, $v'_{perp,B} = \frac{u \cos \theta}{\sin \alpha}$.
   The velocity vector components in the horizontal-vertical frame after collision are:
   For A: $v_{x,A}' = v'_{perp,A} \sin \alpha = -u \cos \theta$.
   $v_{y,A}' = -v'_{perp,A} \cos \alpha = u \cos \theta \cot \alpha$.
   For B: $v_{x,B}' = v'_{perp,B} \sin \alpha = u \cos \theta$.
   $v_{y,B}' = -v'_{perp,B} \cos \alpha = -u \cos \theta \cot \alpha$.
   Let $v_{y,0}' = u \cos \theta \cot \alpha$. So $v_{y,A}' = v_{y,0}'$ and $v_{y,B}' = -v_{y,0}'$.

5. Time of Flight after Collision:

   Let $t_A''$ and $t_B''$ be the times to return to the ground from the collision point $(x_c, y_c)$.
   The vertical motion is $0 = y_c + v_y' t'' - \frac{1}{2} g (t'')^2$.
   The positive root is $t'' = \frac{v_y' + \sqrt{(v_y')^2 + 2 g y_c}}{g}$.
   For A: $t_A'' = \frac{v_{y,0}' + \sqrt{(v_{y,0}')^2 + 2 g y_c}}{g}$.
   For B: $t_B'' = \frac{-v_{y,0}' + \sqrt{(v_{y,0}')^2 + 2 g y_c}}{g}$.
   Subtracting these: $t_A'' - t_B'' = \frac{2 v_{y,0}'}{g} = \frac{2 u \cos \theta \cot \alpha}{g}$. (Eq 4)

6. Using Total Time of Flight:

   Given $T_A = t_A' + t_A'' = 4$ s and $T_B = t_B' + t_B'' = 2$ s.
   Subtracting these: $(t_A' - t_B') + (t_A'' - t_B'') = 4 - 2 = 2$. (Eq 5)

7. Combining Equations:

   From (Eq 3A) and (Eq 3B):
   $t_A' + t_B' = \frac{u}{g \sin \alpha} (\cos(\theta - \alpha) - \cos(\theta + \alpha)) = \frac{u}{g \sin \alpha} (2 \sin \theta \sin \alpha) = \frac{2u \sin \theta}{g}$.
   $t_A' - t_B' = \frac{u}{g \sin \alpha} (\cos(\theta - \alpha) + \cos(\theta + \alpha)) = \frac{u}{g \sin \alpha} (2 \cos \theta \cos \alpha) = \frac{2u \cos \theta \cot \alpha}{g}$.
   Substitute $t_A' - t_B'$ from this into (Eq 5):
   $\frac{2u \cos \theta \cot \alpha}{g} + \frac{2u \cos \theta \cot \alpha}{g} = 2$.
   $2 \left( \frac{2u \cos \theta \cot \alpha}{g} \right) = 2 \implies \frac{2u \cos \theta \cot \alpha}{g} = 1$. (Eq 6)

8. Solving for $\tan \alpha$ and $H_{max}$:

   From (Eq 1): $u \cos \theta (t_A' + t_B') = 24$.
   From (Eq 2): $u \sin \theta = \frac{g}{2} (t_A' + t_B')$.
   Let $t_{sum} = t_A' + t_B'$.
   $u \cos \theta = 24/t_{sum}$.
   $u \sin \theta = g t_{sum}/2$.
   Substitute $u \cos \theta$ into (Eq 6): $\frac{2 (24/t_{sum}) \cot \alpha}{g} = 1 \implies \frac{48 \cot \alpha}{g t_{sum}} = 1 \implies t_{sum} = \frac{48 \cot \alpha}{g}$.
   Equate the two expressions for $t_{sum}$: $\frac{48 \cot \alpha}{g} = \frac{2u \sin \theta}{g} \implies 48 \cot \alpha = 2u \sin \theta$.
   From $u \sin \theta = g t_{sum}/2$:
   $t_{sum} = \frac{48 \cot \alpha}{g} = \frac{48}{10 \tan \alpha}$.
   Now, from $y_c = u \sin \theta t_A' - \frac{1}{2} g (t_A')^2$.
   Also, from $t_A' - t_B' = 1$ (from Eq 5 and Eq 4).
   And $t_A' + t_B' = t_{sum}$.
   So $t_A' = (t_{sum} + 1)/2$ and $t_B' = (t_{sum} - 1)/2$.
   Substitute $t_A' = \frac{1}{2} (\frac{48}{g \tan \alpha} + 1) = \frac{24}{g \tan \alpha} + 0.5$.
   $u \sin \theta = \frac{g}{2} t_{sum} = \frac{g}{2} \frac{48}{g \tan \alpha} = \frac{24}{\tan \alpha}$.
   Substitute these into $y_c = u \sin \theta t_A' - \frac{1}{2} g (t_A')^2$:
   $y_c = \frac{24}{\tan \alpha} \left( \frac{24}{g \tan \alpha} + 0.5 \right) - \frac{1}{2} g \left( \frac{24}{g \tan \alpha} + 0.5 \right)^2$.
   $y_c = \frac{576}{g \tan^2 \alpha} + \frac{12}{\tan \alpha} - \frac{g}{2} \left( \frac{576}{g^2 \tan^2 \alpha} + \frac{24}{g \tan \alpha} + 0.25 \right)$.
   $y_c = \frac{576}{g \tan^2 \alpha} + \frac{12}{\tan \alpha} - \frac{288}{g \tan^2 \alpha} - \frac{12}{\tan \alpha} - \frac{g}{8}$.
   $y_c = \frac{288}{g \tan^2 \alpha} - \frac{g}{8}$.
   Now use $t_A'' = \frac{v_{y,0}' + \sqrt{(v_{y,0}')^2 + 2 g y_c}}{g}$.
   And $t_A'' = 4 - t_A' = 4 - (\frac{24}{g \tan \alpha} + 0.5) = 3.5 - \frac{24}{g \tan \alpha}$.
   Substitute $v_{y,0}' = g/2$ (from Eq 6) and $y_c$:
   $3.5 - \frac{24}{g \tan \alpha} = \frac{g/2 + \sqrt{(g/2)^2 + 2g(\frac{288}{g \tan^2 \alpha} - \frac{g}{8})}}{g}$.
   $3.5 - \frac{24}{g \tan \alpha} = \frac{1}{2} + \frac{\sqrt{g^2/4 + \frac{576}{\tan^2 \alpha} - g^2/4}}{g}$.
   $3.5 - \frac{24}{g \tan \alpha} = \frac{1}{2} + \frac{\sqrt{576/\tan^2 \alpha}}{g}$.
   $3.5 - \frac{24}{g \tan \alpha} = \frac{1}{2} + \frac{24}{g \tan \alpha}$.
   $3 = \frac{48}{g \tan \alpha}$.
   $\tan \alpha = \frac{48}{3g} = \frac{16}{g} = \frac{16}{10} = \frac{8}{5}$.
   So, option (B) is correct.

9. Calculate Maximum Height:

   Maximum height $H_{max} = \frac{(u \sin \theta)^2}{2g}$.
   We have $u \sin \theta = \frac{24}{\tan \alpha} = \frac{24}{8/5} = 24 \times \frac{5}{8} = 3 \times 5 = 15$ m/s.
   $H_{max} = \frac{15^2}{2 \times 10} = \frac{225}{20} = 11.25$ m.
   So, option (A) is correct.

The final answer is $\boxed{AB}$