Question

A projectile is projected with velocity of 25 m/s at an angle θ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of θ will be : [use g = 10 m/s² ]

(A) \(\frac12sin^{-1}(5t^2/4R)\) 

(B) \(\frac12sin^{-1}(4R/5t^2)\) 

(C) tan^-1(4t²/5R)

(D) cot^-1(R/20t²)

Answer 1

Correct option is (D) \( \cot^{-1}\left[\frac R{20t^2}\right]\)

\(t = \frac{25 \sin \theta}g\)

and, \(R = \frac{(25)^2 (2\sin\theta\cos \theta)}{g}\)

⇒ \(R = \frac{25 \times 25 \times 2}{g} \times \frac{gt}{25} \times \cos \theta \)

⇒ \(R = 50t \cos \theta\)

\(\therefore \) \(\tan \theta =\frac{gt}{25} \times \frac{50t}R\)

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

⇒ \(\theta = \cot^{-1}\left[\frac R{20t^2}\right]\)

Answer 2

Correct option is (D) cot^-1(R/20t²)

 \(\theta = \cot^{-1} \left(\frac{R}{20t^2}\right)\)