A particle $A$ is projected vertically upwards. Another identical particle $B$ is projected at an angle of $45^o $ . Both reach the same height. The ratio of the initial kinetic energy of $A$ to that of $B$ is
A particle $A$ is projected vertically upwards. Another identical particle $B$ is projected at an angle of $45^o $ . Both reach the same height. The ratio of the initial kinetic energy of $A$ to that of $B$ is
- A
$1:2$
- B
$2:1$
- C
$1:\sqrt 2$
- D
$\sqrt 2:1$
Similar Questions
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$\theta_{0}=\tan ^{-1}\left(\frac{4 h_{m}}{R}\right)$
Where the symbols have their usual meaning.
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$\theta(t)=\tan ^{-1}\left(\frac{v_{0 y}-g t}{v_{0 x}}\right)$
Show that the projection angle $\theta_{0}$ for a projectile launched from the origin is given by
$\theta_{0}=\tan ^{-1}\left(\frac{4 h_{m}}{R}\right)$
Where the symbols have their usual meaning.
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