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Question

Calculate the projection angle of projectile whose range is $$ 50m $$ and maximum height $$ 10 m $$ 


Solution

Given $$ R = 50 m ; A = 10 m ; \theta  =  ? $$
$$ \because Range  \quad R = \frac {u^2 sin  2 \theta}{g} $$
And ,maximum height
$$ H = \frac {u^2 sin^2 \theta}{2g} $$
$$ \therefore \frac {R}{H} = \frac  { \frac { u^2 sin 2 \theta}{g} }{ \frac { u^2 sin^2 \theta}{2g} } = \frac { 2 sin  2 \theta}{ sin^2 \theta } = \frac { 2 \times  2 sin  \theta cos  \theta }{ sin^2 \theta } $$
or $$ \frac {50}{10} = 4 \frac {cos  \theta }{ sin  \theta } = 4 cot  \theta  = \frac {4}{tan  \theta } $$
or $$ \frac {5}{1} = \frac { 4}{ tan \theta } = 4 cot \theta  = \frac {4}{ tan \theta } $$
or $$  \frac {5}{1} = \frac {4}{ tan \theta } \Rightarrow tan  \theta  = \frac {4}{5} = 0.8 $$
$$ \therefore \theta = tan^{-1} (0.8) = 38.66^0 $$
$$ \theta  = 38.66^0 $$

Physics

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