Question

A ball is projected with same velocity at an angle Theta and (90-Theta) with horizontal. If h1 and h2 are maximum heights attained by it in two paths and R is the range of projectile, then which of the following relation is correct?

(a) R=h1h2 (b) R=h1+h2
(c) R=4(h1h2)^1/2 (c) R=(h1h2)^1/3

Answer 1

the correct answer is c)R=4(h1h2)^1/2

Answer 2

The correct relation is (c) $R\quad =\quad 4\sqrt { h_{ 1 }h_{ 2 } }$

Where, $h\quad =\quad \frac { u^{ 2 }\sin^{ 2 } \theta }{ 2g }$

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

           $\frac { h }{ R } \quad =\quad \frac { u^{ 2 }\sin ^{ 2 } \theta }{ 2g } \quad \times \quad \frac { g }{ u^{ 2 }\sin 2\theta }$

           $\because \quad \sin 2\theta \quad =\quad 2\sin \theta \cos \theta$

           $\frac { h }{ R } \quad =\quad \frac { u^{ 2 }\sin ^{ 2 } \theta }{ 2g } \quad \times \quad \frac { g }{ u^{ 2 }(2\sin \theta \cos \theta ) }$

On cancelling, we get,

           $\frac { h }{ R } \quad =\quad \frac { \tan \theta }{ 4 }$

(By substituting the values of h and R we get the above equation)

           $h\quad =\quad \frac { R\tan \theta }{ 4 }$

           $h_{ 1 }\quad =\quad \frac { R\tan \theta }{ 4 }$

           $h_{ 2 }\quad =\quad \frac { R\tan \left( 90°\quad -\quad \theta \right) }{ 4 } \quad =\quad R\frac { \cot \theta }{ 4 }$

           $h_{ 1 }h_{ 2 }\quad =\quad \frac { R^{ 2 } }{ 16 }$

           $\because \quad \tan  \theta \cot  \theta \quad =\quad 1$

           $16h_{ 1 }h_{ 2 }\quad =\quad R^{ 2 }$

           $R\quad =\quad 4\sqrt { h_{ 1 }h_{ 2 } }$