Skip to main content

Visit this link for 1 : 1 LIVE Classes.

The ranges and heights for two projectiles projected ....

The ranges and heights for two projectiles projected with the same initial velocity at angles $42^\circ $ and $48^\circ $ with the horizontal are $R_1\,,R_2$ and $H_1\,,H_2$ respectively. Choose the correct option:

(A) $R_1 < R_2 $ and $H_1 < H_2 $
(B) $R_1 > R_2 $ and $H_1 = H_2 $
(C) $R_1 = R_2 $ and $H_1 = H_2 $
(D) $R_1 = R_2 $ and $H_1 < H_2 $

Solution

$R=\frac {u^2 sin 2\theta }{g} $, $H=\frac {u^2 sin^2 \theta }{2g}$ 

$\frac {R_1}{R_2} = \frac {sin 2 \times 42^\circ}{sin 2\times 48^\circ} = \frac {cos 6^\circ}{cos 6^\circ} = 1 $

$\frac {H_1}{H_2}= \frac {sin^2 42^\circ }{sin^2 48^\circ } = tan^2 42^\circ < 1 $

Answer: (D)

Popular posts from this blog

${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)