Question

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

• A. $\frac{1}{2}sin^{-1]}[\frac{5t^2}{4R}]$
• B. $\frac{1}{2}sin^{-1]}[\frac{4R}{5t^2}]$
• C. tan-1[$\frac{4t^2}{4R}$]
• D. cot-1[$\frac{R}{20t^2}$]

Answer 1

Answer: (D)

cot-1[$\frac{R}{20t^2}$]

Answer 2

t=$\frac{25sinθ}{g}$
and, R=$\frac{(25)^2(2sinθcosθ)}{g}$

$\Rightarrow \text{} R = \frac{25\times25\times2}{g}\times\frac{gt}{25}\times cos \theta$
$\Rightarrow R = 50t cos \theta$

$\therefore tan \theta = \frac{gt}{25}\times \frac{50t}{R}$
$\Rightarrow tan \theta =\frac{20t^2}{R}$

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

The correct option is (D): cot-1[$\frac{R}{20t^2}$]