Question

A cannon shoots an artillery shell with an 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

It will land at 14139.19 m away.

Explanation:

The expression for range d on level ground is given by;

d=v² sin (2Ф) /g where Ф is the fire angle and g is acceleration due to gravity

Given v=400m/s ,Ф= 60° and g=9.8 so,

d= 400² sin(120°) /9.8

d=(400²×0.86602540378) / 9.8

d=14139.19 m

Motion for falling object : https://brainly.com/question/11799308

Keyword : initial velocity, angle, range

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