Question

If R and H represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is

Answer 1

Answer:

$\alpha = { \tan }^{ - 1} \frac{4h}{ \: r}$

Explanation:

Given---> r and h are horizontal range and maximum height of the projectile .

To find -----> Angle of projection

Concept used ----> If angle of projection and initial velocity of projectile be

$\alpha and \: u$

then ,

$r \: = \frac{ {u}^{2} \: \sin(2 \alpha ) }{g}$

and

$h \: = \frac{ {u}^{2} \: { \sin }^{2} \alpha }{2g}$

Solution-----> Now we know that ,

$r \: = \frac{ {u}^{2} \sin(2 \alpha ) }{g}$

$h \: = \frac{ {u}^{2} { \sin }^{2} \alpha }{2g}$

$now \: dividing \: first \: relation \: by \: second \: one \: we \: get$

$\frac{r}{h} = \frac{ \frac{ {u}^{2} \: \sin(2 \alpha ) }{g} }{ \frac{ {u}^{2} \: { \sin \ }^{2} \alpha }{2g} }$

${u}^{2} and \: g \: cancel \: out \: from \: the \: numerator \:and \: denominator$

$\frac{r}{h} = \frac{2 \: \sin(2 \alpha ) }{ \ \ { \sin }^{2} \alpha }$

$we \: know \: that \:$

$\sin(2 \alpha ) = 2 \: \sin( \alpha ) \cos( \alpha )$

$applying \: it \: we \: get$

$\frac{r}{h} = \frac{2 \: \times 2 \: \sin \alpha \: \cos \alpha }{ \ { \sin }^{2} \alpha }$

$\sin \alpha \: cancel \: out \: from \: numerator \: and \: denominator$

$\frac{r}{h} = 4 \: \cot \alpha$

$\cot \alpha = \frac{r}{4h}$

$\tan \alpha = \frac{4h}{r}$

$so$

$\alpha = \ { \tan}^{ - 1} ( \: \frac{4h}{r} )$