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Motion In A Plane

Question
CBSEENPH11018588

Show that if the projectile attains the maximum height of H and falls at a distance R, then it is projected at an angle θ given by <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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Solution
Let the speed with which the projectile be projected at angle θ  be 'u'.
The maximum height and the horizontal range is given by,
Horizontal range, straight R equals fraction numerator straight u squared sin 2 straight theta over denominator straight g end fraction
Maximum height, straight H equals fraction numerator straight u squared sin squared straight theta over denominator 2 straight g end fraction 
Ratio of horizontal range to maximum height is given by, 
∴               straight R over straight H equals fraction numerator 2 s i n 2 straight theta over denominator s i n squared straight theta end fraction equals 4 space c o t space straight theta
rightwards double arrow          t a n space straight theta space equals space fraction numerator 4 straight H over denominator straight R end fraction
That is, 
                  straight theta equals t a n to the power of negative 1 end exponent open parentheses fraction numerator 4 straight H over denominator straight R end fraction close parentheses, is the angle of projection.