JEE MAIN - Physics (2022 - 24th June Morning Shift - No. 5)
A projectile is projected with velocity of 25 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 m/s2]
$${1 \over 2}{\sin ^{ - 1}}\left( {{{5{t^2}} \over {4R}}} \right)$$
$${1 \over 2}{\sin ^{ - 1}}\left( {{{4R} \over {5{t^2}}}} \right)$$
$${\tan ^{ - 1}}\left( {{{4{t^2}} \over {5R}}} \right)$$
$${\cot ^{ - 1}}\left( {{R \over {20{t^2}}}} \right)$$
Explanation
_24th_June_Morning_Shift_en_5_1.png)
$$t = {{25\sin \theta } \over g}$$
and, $$R = {{{{(25)}^2}(2\sin \theta \cos \theta )} \over g}$$
$$ \Rightarrow R = {{25 \times 25 \times 2} \over g} \times {{gt} \over {25}} \times \cos \theta $$
$$ \Rightarrow R = 50t\cos \theta $$
$$\therefore$$ $$tan\theta = {{gt} \over {25}} \times {{50t} \over R}$$
$$ = {{20{t^2}} \over R}$$
$$ \Rightarrow \theta = {\cot ^{ - 1}}\left( {{R \over {20{t^2}}}} \right)$$
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