Question

A particle is thrown with a speed \(v\) from a point \(O\) at an angle \(\theta\) with the horizontal plane such that it passes through the point \(P\) at a height of \(1\,\text{m}\) and horizontal distance of \(5\,\text{m}\) from \(O\), as shown in the figure. If acceleration due to gravity is \(g\,\text{m s}^{-2}\), then the correct statement(s) is/are:

• A. If \(\theta=45^\circ\), then \[ v=\frac{5\sqrt g}{2}\,\mathrm{m\,s^{-1}} \]
• B. If \(\theta=45^\circ\), the particle reaches its maximum height before it reaches \(P\)
• C. If \(\theta=30^\circ\), the particle reaches its maximum height after reaching \(P\)
• D. If \(\theta=\tan^{-1}\left(\frac15\right)\), then \[ v=125\sqrt g\,\mathrm{m\,s^{-1}} \]

Hint:

Projectile trajectory: \[ y=x\tan\theta-\frac{gx^2}{2v^2\cos^2\theta} \] Horizontal coordinate of maximum height: \[ x_H=\frac{v^2\sin2\theta}{2g} \]

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem deals with projectile motion. The trajectory of a particle launched from the origin is given by a parabolic equation relating the vertical coordinate \( y \) to the horizontal coordinate \( x \). We are given a specific point \( (x, y) = (5, 1) \) that the projectile must pass through. By substituting these coordinates and the given angles into the trajectory equation, we can determine the required initial velocity and the location of the peak of the path.
Step 2: Key Formula or Approach:
The equation of trajectory is:
\[ y = x \tan \theta - \frac{g x^2}{2 v^2 \cos^2 \theta} \]
Maximum height is reached at a horizontal distance \( X_h = \frac{R}{2} = \frac{v^2 \sin(2\theta)}{2g} \).
Step 3: Detailed Explanation:
Let's check statement (A): Put \( x=5, y=1, \theta=45^{\circ} \).
\[ 1 = 5 \tan 45^{\circ} - \frac{g(5)^2}{2 v^2 \cos^2 45^{\circ}} \]
\[ 1 = 5(1) - \frac{25 g}{2 v^2 (1/2)} \implies 1 = 5 - \frac{25 g}{v^2} \]
\[ \frac{25 g}{v^2} = 4 \implies v^2 = \frac{25 g}{4} \implies v = \frac{5\sqrt{g}}{2} \]
So, (A) is correct.
Let's check statement (B): For \( \theta=45^{\circ} \), we found \( v^2 = \frac{25g}{4} \).
The horizontal distance to the maximum height is:
\[ X_h = \frac{v^2 \sin(90^{\circ})}{2g} = \frac{(25g/4) \cdot 1}{2g} = \frac{25}{8} = 3.125 \text{ m} \]
The horizontal distance of point \( P \) is 5 m. Since \( 3.125<5 \), the particle reaches its peak before reaching \( P \).
So, (B) is correct.
Let's check statement (C): Put \( x=5, y=1, \theta=30^{\circ} \).
\[ 1 = 5 \tan 30^{\circ} - \frac{g(25)}{2 v^2 \cos^2 30^{\circ}} = \frac{5}{\sqrt{3}} - \frac{25 g}{2 v^2 (3/4)} \]
\[ \frac{50 g}{3 v^2} = \frac{5}{\sqrt{3}} - 1 \implies v^2 = \frac{50 g}{3 (5/\sqrt{3} - 1)} \approx 8.8g \]
Horizontal distance to peak: \( X_h = \frac{v^2 \sin 60^{\circ}}{2g} = \frac{8.8g \cdot \sqrt{3}/2}{2g} \approx 3.8 \text{ m} \).
Again, \( 3.8<5 \), so the peak is reached before \( P \). Thus, (C) is incorrect.
Let's check statement (D): If \( \tan \theta = 1/5 \), then the line joining \( O \) to \( P \) has a slope of \( 1/5 \). This means the particle is launched exactly towards \( P \). Due to gravity, the particle must always fall below this straight line. Therefore, it can only pass through \( P \) if gravity is zero or if it has infinite velocity. Mathematically:
\[ 1 = 5(1/5) - \frac{gx^2}{2v^2\cos^2\theta} \implies 1 = 1 - \text{positive term} \]
This equation has no real solution for \( v \). Thus, (D) is incorrect.
Step 4: Final Answer:
The particle must be launched with \( v = 2.5\sqrt{g} \) at \( 45^{\circ} \) to hit \( P \), and at that speed, the peak occurs at \( x=3.125 \text{ m} \). Launching at the same angle as the target point's position vector is impossible for any finite velocity.