Acannon 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

14139.19 m

Explanation:

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

(1)

(2)

Where:

is the horizontal distance (where the artillery shell lands)

is the initial velocity

is the angle

is the time

is the final height

is the initial height

is the acceleration due gravity, always directed downwards

So, let's begin by isolating from (2):

(3)

(4)

Substituting (4) in (1):

(5)

Rewriting (5) and taking into account :

(6)

(7)

Finally:

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.