Question

Bullet is fired from the gun it covers maximum horizontal distance of 10 kilometres determine the velocity of projection and the maximum height attained by the bullet​

Answer 1

Answer:

The velocity of projection = 313.05 m/s

Height of projection = 2500 m

Explanation:

If the bullet covers maximum horizontal distance then the angle of projection must be 45°

Given the maximum range = 10 km = 10000 m

If the initial velocity is u then

$R_{max}=\frac{u^2}{g}$

or $u^2= gR_{max}$

$\implies u^2=  9.8\times10000$

$\implies u^2=98000$

$\implies u=313.05$ m/s

Maximum height

$H=\frac{u^2\sin^245}{2g}$

$\implies H=\frac{98000\times(1/2)}{2 \times9.8}$

or, H = 2500 m