Physics

Problem

A projectile is projected with velocity of

25 \text{ m/s}

at an angle

\theta

with the horizontal. After

t

seconds its inclination with horizontal becomes zero. If

R

represents horizontal range of the projectile, the value of

\theta

will be: [use

g = 10 \text{ m/s}^2

] [Main June 24, 2022 (II)] (a)

\frac{1}{2} \sin^{-1} \left( \frac{5t^2}{4R} \right)

(b)

\frac{1}{2} \sin^{-1} \left( \frac{4R}{5t^2} \right)

(c)

\tan^{-1} \left( \frac{4t^2}{5R} \right)

(d)

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