Question 79: Motion in 2D & Vectors

Question

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

Accepted Answer

$$t = {{25\sin 	heta } \over g}$$
and, $$R = {{{{(25)}^2}(2\sin 	heta \cos 	heta )} \over g}$$
$$ \Rightarrow R = {{25 	imes 25 	imes 2} \over g} 	imes {{gt} \over {25}} 	imes \cos 	heta $$
$$ \Rightarrow R = 50t\cos 	heta $$
$$	herefore$$ $$tan	heta  = {{gt} \over {25}} 	imes {{50t} \over R}$$
$$ = {{20{t^2}} \over R}$$
$$ \Rightarrow 	heta  = {\cot ^{ - 1}}\left( {{R \over {20{t^2}}}} ight)$$

Answer

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

Answer

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

Answer

$${	an ^{ - 1}}\left( {{{4{t^2}} \over {5R}}} ight)$$

Answer

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

Motion in 2D & Vectors

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]

Choose the correct answer: