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

Step-by-step 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+(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+(g)/(2)t^(2)` (3)

`t=-(2 V_(o)sin \theta)/(g)` (4)

Substituting (4) in (1):

`x=V_(o)cos \theta (-(2 V_(o)sin \theta)/(g))` (5)

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

`x=-(V_(o)^(2)sin(2\theta))/(g)` (6)

`x=-((400 m/s)^(2)sin(2(60\°)))/(-9.8 m/s^(2))` (7)

Finally:

`x=14139.19 m`