Question

A projectile is projected with velocity of \( 40 \) m/s at an angle \( \theta \) with the horizontal. If \( R \) is the horizontal range covered by the projectile and after \( t \) seconds its inclination with horizontal becomes zero, then the value of \( \cot \theta \) is:
[Take, \( g = 10 \) m/s\( ^2 \)]

• A. \( \frac{R}{20t^2} \)
• B. \( \frac{R}{10t^2} \)
• C. \( \frac{5R}{t^2} \)
• D. \( \frac{R}{t^2} \)

Hint:

For projectile motion, the horizontal range is given by: \[ R = \frac{u^2 \sin 2\theta}{g} \] Using this formula helps in solving range-related problems quickly.

Answer 1

The Correct Option is A

Step 1: {Find the time to reach maximum height}
At maximum height, the inclination with the horizontal becomes zero.
The time to reach maximum height is: \[ t = \frac{u \sin \theta}{g} \] Rearranging: \[ u = \frac{g t}{\sin \theta} \] Step 2: {Use the range formula}
\[ R = u \cos \theta \times (2t) \] Substituting \( u \): \[ R = \frac{g t}{\sin \theta} \cos \theta \times (2t) \] \[ \cos \theta = \frac{R}{2 u t} \] Step 3: {Find \( \cot \theta \)}
\[ \cot \theta = \frac{R}{2 g t^2} \] Substituting \( g = 10 \): \[ \cot \theta = \frac{R}{2 \times 10 t^2} = \frac{R}{20 t^2} \] Thus, the correct answer is (A) \( \frac{R}{20t^2} \).

Answer 2

Given:
- Initial speed \( u = 40 \) m/s
- Angle of projection = \( \theta \)
- Range = \( R \)
- After time \( t \), the projectile's inclination with the horizontal becomes zero (i.e., vertical velocity = 0)
- Take \( g = 10 \, \text{m/s}^2 \)

Step 1: Understand the condition
The condition that the inclination with the horizontal becomes zero after time \( t \) implies:
- The **vertical component of velocity** becomes zero at that time
- So, the projectile is at **maximum height** at time \( t \)

Time to reach maximum height is:
\[ t = \frac{u \sin \theta}{g} \Rightarrow \sin \theta = \frac{g t}{u} = \frac{10t}{40} = \frac{t}{4} \]

Step 2: Use range formula
The range \( R \) is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Use identity: \( \sin 2\theta = 2 \sin \theta \cos \theta \)
\[ R = \frac{u^2 \cdot 2 \sin \theta \cos \theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g} \]

Substitute \( \sin \theta = \frac{t}{4} \):
\[ R = \frac{2u^2}{g} \cdot \frac{t}{4} \cdot \cos \theta \Rightarrow R = \frac{2 \cdot 1600}{10} \cdot \frac{t}{4} \cdot \cos \theta = 320 \cdot \frac{t}{4} \cdot \cos \theta = 80t \cos \theta \]

Step 3: Express \( \cot \theta \)
We know \( \cot \theta = \frac{\cos \theta}{\sin \theta} \)
From earlier, \( \sin \theta = \frac{t}{4} \), and we now have \( \cos \theta = \frac{R}{80t} \)
So:
\[ \cot \theta = \frac{R}{80t} \div \frac{t}{4} = \frac{R}{80t} \cdot \frac{4}{t} = \frac{R \cdot 4}{80t^2} = \frac{R}{20t^2} \]

Final Answer:
\[ \boxed{\cot \theta = \frac{R}{20t^2}} \]