Question

A person observes a bird on a tree 39.6 m at a distance of 59.2m with what velocity the
person should throw an arrow at an angle of 45 degree. So that it may hit the bird? ​

Answer 1

Given:

A person observes a bird on a tree 39.6 m at a distance of 59.2m

To find:

Velocity the person should throw an arrow at an angle of 45 degree to hit the target.

Calculation:

Let Velocity be v ;

Along X axis:

$x = v \cos( \theta) t$

$= > 59.2 = v \cos( {45}^{ \circ} ) t$

Along Y axis:

$y = v \sin( \theta) t - \dfrac{1}{2} g {t}^{2}$

$= > y = v \sin( {45}^{ \circ} ) t - \dfrac{1}{2} g {t}^{2}$

Putting value of t ;

$= > 39.6= v \sin( {45}^{ \circ} ) \{\dfrac{59.2}{v \cos( {45}^{ \circ} ) } \} - \dfrac{1}{2} g {t}^{2}$

$= > 39.6= 59.2\tan( {45}^{ \circ}) - \dfrac{1}{2} g { \{ \dfrac{59.2}{v \cos( {45}^{ \circ} ) } \} }^{2}$

$= > 39.6= 59.2 - \dfrac{1}{2} g { \{ \dfrac{59.2}{v \cos( {45}^{ \circ} ) } \} }^{2}$

$= > \dfrac{1}{2} g { \{ \dfrac{59.2}{v \cos( {45}^{ \circ} ) } \} }^{2} = 19.6$

$= > \dfrac{1}{2} (9.8) { \{ \dfrac{59.2}{v \cos( {45}^{ \circ} ) } \} }^{2} = 19.6$

$= > \dfrac{1}{2} { \{ \dfrac{59.2}{v \cos( {45}^{ \circ} ) } \} }^{2} = 2$

$= > { \{ \dfrac{59.2}{v \cos( {45}^{ \circ} ) } \} }^{2} = 4$

$= > \{ \dfrac{59.2}{v \cos( {45}^{ \circ} ) } \} = 2$

$= > v = \{ \dfrac{59.2}{2 \cos( {45}^{ \circ} ) } \}$

$= > v = \dfrac{59.2}{2 \times \frac{1}{ \sqrt{2} } }$

$= > v = \dfrac{59.2}{ \sqrt{2} }$

$= > v = 41.86 \: m {s}^{ - 1}$

So, final answer is:

$\boxed{ \sf{v = 41.86 \: m {s}^{ - 1} }}$