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A stone projected with a velocity u at an angle $\theta$ with the horizontal reaches maximum height $H_1$. When it is projected with velocity u at an angle $\left( {\frac{\pi }{2} - \theta } \right)$ with the horizontal, it reaches maximum height $ H_2$. The relation between the horizontal range R of the projectile, $H_1$ and $H_2$ is
$R = 4\sqrt {{H_1}{H_2}} $
$R = 4({H_1} - {H_2})$
$R = 4({H_1} + {H_2})$
$R = \frac{{{H_1}^2}}{{{H_2}^2}}$
Solution
(a)${H_1} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ and ${H_2} = \frac{{{u^2}{{\sin }^2}(90 – \theta )}}{{2g}} = \frac{{{u^2}{{\cos }^2}\theta }}{{2g}}$
${H_1}{H_2} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}} \times \frac{{{u^2}{{\cos }^2}\theta }}{{2g}} = \frac{{{{({u^2}\sin 2\theta )}^2}}}{{16{g^2}}} = \frac{{{R^2}}}{{16}}$
$R = 4\sqrt {{H_1}{H_2}} $
