Question

An artillery shell is fired at an angle of 26.8◦ above the horizontal ground with an initial speed of 1660 m/s.
Find the total time of flight of the shell, neglecting air resistance.

Answer 1

Task:

An artillery shell is fired at an angle of 26.8° above the horizontal ground with an initial speed of 1660 m/s.

Find the total time of flight of the shell, neglecting air resistance.

Solution:

We have a coordinate system where an axis $Oy$ is perpendicular to the ground. This coordinate system is in the plane of flight of the shell. So, it's initial speed consists of two components: $v_x$ and $v_y$:

$$v_x = v \cos \alpha$$ $$v_y = v \sin \alpha$$

When the half time of flight of the shell will pass, the speed $v_y$ will become slower because of the action of the gravity force and its value will become zero:

$$v_y' = v_y - \frac{gt}{2}$$ $$v_y' = 0$$ $$v_y = \frac{gt}{2}$$

So, the total time of flight of the shell will be:

$$t = \frac{2v \sin \alpha}{g}$$ $$t = \frac{2 \cdot 1660 \, \mathrm{m/s} \cdot \sin 26.8{}^\circ}{9.8 \, \mathrm{m/s}^2} = 152.75 \, \mathrm{s}$$

THE ANSWER: t=152.75 s.