Question

The initial kinetic energy imparted to a 0.022 bullet is 1300 J. Find the range of this projectile when it is fired at an angle such that the range is equals the maximum height attained.

Answer 1

$l_{max}=v^2\sin(2\alpha)/g$

$h_{max}=v^2\sin^2\alpha/(2g)$

$v^2\sin^2\alpha/(2g)=v^2\sin(2\alpha)/g\to\alpha=75.96°$

$mv^2/2=1300\to v=\sqrt{2600/0.022}=344\ (m/s)$. So,

$l_{max}=v^2\sin(2\alpha)/g=344^2\cdot\sin(2\cdot75.96)/9.8=5684\ (m)$ . Answer