Question

A projectile is thrown at a speed V and at an angle with the horizontal. If the speed at its maximum height is \(\frac{V}{3}\),then the value of tan θ is: 

• A. \(\sqrt{3}\)
• B. \(\frac{1}{\sqrt{3}}\)
• C. 2\(\sqrt{2}\)
• D. 3
• E. 3\(\sqrt{3}\)

Answer 1

The Correct Option is C

Given:

• Initial speed of projectile: \( V \)
• Launch angle: \( \theta \)
• Speed at maximum height: \( \frac{V}{3} \)

Step 1: Analyze Velocity Components

At maximum height:

• Vertical velocity component (\( v_y \)) becomes 0.
• Horizontal velocity component (\( v_x \)) remains constant throughout the flight.

Thus, the speed at maximum height equals the horizontal component:

\[ v_x = \frac{V}{3} \]

Step 2: Relate to Initial Velocity

The horizontal component of the initial velocity is:

\[ v_x = V \cos \theta \]

From Step 1:

\[ V \cos \theta = \frac{V}{3} \]

Simplify:

\[ \cos \theta = \frac{1}{3} \]

Step 3: Find \(\tan \theta\)

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

\[ \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \]

Now, calculate \( \tan \theta \):

\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{2\sqrt{2}}{3}}{\frac{1}{3}} = 2\sqrt{2} \]

Conclusion:

The value of \( \tan \theta \) is \( 2\sqrt{2} \).

Answer: \(\boxed{C}\)

Answer 2

Step 1: Recall the components of velocity in projectile motion.

In projectile motion, the velocity can be resolved into horizontal and vertical components:

• The horizontal component of velocity is \( V_x = V \cos\theta \), which remains constant throughout the motion because there is no horizontal acceleration.
• The vertical component of velocity is \( V_y = V \sin\theta \), which changes due to gravity.

 

At the maximum height, the vertical component of velocity becomes zero (\( V_y = 0 \)), and the total speed is equal to the horizontal component of velocity: \[ \text{Speed at maximum height} = V_x = V \cos\theta. \]

We are given that the speed at maximum height is \( \frac{V}{3} \). Therefore: \[ V \cos\theta = \frac{V}{3}. \]

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Step 2: Solve for \( \cos\theta \).

Divide both sides of the equation by \( V \) (assuming \( V \neq 0 \)): \[ \cos\theta = \frac{1}{3}. \]

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Step 3: Use the trigonometric identity to find \( \tan\theta \).

The relationship between \( \sin\theta \), \( \cos\theta \), and \( \tan\theta \) is given by: \[ \tan\theta = \frac{\sin\theta}{\cos\theta}. \]

Using the Pythagorean identity \( \sin^2\theta + \cos^2\theta = 1 \), we can solve for \( \sin\theta \): \[ \sin^2\theta = 1 - \cos^2\theta. \] Substitute \( \cos\theta = \frac{1}{3} \): \[ \sin^2\theta = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9}. \] Take the square root: \[ \sin\theta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}. \] Now calculate \( \tan\theta \): \[ \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{2\sqrt{2}}{3}}{\frac{1}{3}} = 2\sqrt{2}. \] ---

Final Answer: The value of \( \tan\theta \) is \( \mathbf{2\sqrt{2}} \), which corresponds to option \( \mathbf{(C)} \).