Question

A bullet of fire for a gun with muzzle velocity of 200 metre per second of angle of 60 to the horizontal find the range and maximum height

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

u = 200 m/s.

${ \theta = 60 \degree}$

$\huge{\bold{\underline{Explanation:-}}}$

Case-1

Range is the maximum distance covered in horizontal direction.

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

Substituting the values.

$\large{R = \frac{(200)^2 \sin 2 \times 60}{g}}$

$\large{R = \frac{40000 \: \sin 120}{10}}$

$\large{R = 4000 \times \frac{ \sqrt{3} }{2} }$

$\large{R = { \cancel{4000} \times \frac{ \sqrt{3}}{ \cancel{2}}}}$

$\large{R = 2000 \sqrt{3} \: m}$

$\huge{\boxed{\boxed{R = 2000 \sqrt{3} \: m}}}$

Case-2

Maximum Height is the maximum distance covered in vertical direction.

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

Substituting the values.

$\large{R = \frac{(200)^2 \sin 60}{2 \times 10}}$

$\large{R = \frac{40000 \: \sin 60}{20}}$

$\large{R = {2000 \times \frac{ \sqrt{3}}{2}}}$

$\large{R = { \cancel{2000} \times \frac{ \sqrt{3}}{ \cancel{2}}}}$

$\large{R = 1000 \sqrt{3} \: m}$

$\huge{\boxed{\boxed{R = 1000 \sqrt{3} \: m}}}$

So,the Range is 2000√3 meters and Maximum height is 1000√3 meters .