Question:medium

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\( ^2 \)]

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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.
Updated On: Jan 13, 2026
  • \( \frac{R}{20t^2} \)
  • \( \frac{R}{10t^2} \)
  • \( \frac{5R}{t^2} \)
  • \( \frac{R}{t^2} \)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: {Calculate the time to attain maximum altitude}
At maximum altitude, the projectile's vertical velocity component is zero. The time to reach this point is calculated as: \[t = \frac{u \sin \theta}{g}\] This equation can be rearranged to express \(u\) in terms of \(g\), \(t\), and \(\theta\): \[u = \frac{g t}{\sin \theta}\] Step 2: {Apply the horizontal range equation}
The horizontal range (\(R\)) of a projectile is given by: \[R = u \cos \theta \times (2t)\] Substitute the expression for \(u\) derived in Step 1: \[R = \frac{g t}{\sin \theta} \cos \theta \times (2t)\] Rearrange to isolate \(\cos \theta\): \[\cos \theta = \frac{R}{2 u t}\] Step 3: {Determine \( \cot \theta \)}
Using the derived relationships, we can express \(\cot \theta\) as: \[\cot \theta = \frac{R}{2 g t^2}\] Given that \(g = 10\): \[\cot \theta = \frac{R}{2 \times 10 t^2} = \frac{R}{20 t^2}\] Therefore, the correct option is (A) \( \frac{R}{20t^2} \).
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