Question

A cannon shoots an artillery shell with a initial velocity of 400 meters/second at an indirect fire angle of 60 on level ground where does it land if friction effects are neglected

Answer 1

Answer: 14139.19 m

Explanation:

This situation is related to parabolic motion and can be solved using the following equations:

$x=V_{o}cos \theta t$ (1)

$y=y_{o}+V_{o} sin \theta t+\frac{g}{2}t^{2}$ (2)

Where:

$x$ is the horizontal distance (where the artillery shell lands)

$V_{o}=400 m/s$ is the initial velocity

$\theta=60\°$ is the angle

$t$ is the time

$y=0 m$ is the final height

$y_{o}=0 m$ is the initial height

$g=-9.8 m/s^{2}$ is the acceleration due gravity, always directed downwards

So, let's begin by isolating $t$ from (2):

$0=V_{o} sin \theta t+\frac{g}{2}t^{2}$ (3)

$t=-\frac{2 V_{o}sin \theta}{g}$ (4)

Substituting (4) in (1):

$x=V_{o}cos \theta (-\frac{2 V_{o}sin \theta}{g})$ (5)

Rewriting (5) and taking into account $sin(2\theta)=2 sin \theta cos \theta$:

$x=-\frac{V_{o}^{2}sin(2\theta)}{g}$ (6)

$x=-\frac{(400 m/s)^{2}sin(2(60\°))}{-9.8 m/s^{2}}$ (7)

Finally:

$x=14139.19 m$