Answer:
[tex]2.49\cdot 10^5 m[/tex]
Explanation:
The horizontal range of a projectile is given by
[tex]d=\frac{u^2 sin 2\theta}{g}[/tex]
where
u is the initial speed
[tex]\theta[/tex] is the angle of launch
g = 9.8 m/s^2 is the acceleration of gravity
For the shell in this problem,
u = 1570 m/s
[tex]\theta=41^{\circ}[/tex]
Substituting into the equation, we find the range of the shell:
[tex]d=\frac{(1570)^2}{9.8} sin (2\cdot 41)=2.49\cdot 10^5 m[/tex]
