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) \( \frac{R}{20t^2} \).