Bullet fired w certain velocity at angle above the horizontal at location where g is 10 m a sec2 The initial x and y comp. of its velocity are 86.6 ms and 50.0 ms how long to reach highest trajectory?

Answer 1

The question is already resolved into horizontal and vertical components for us.

This is a very thoughtful and considerate favor, because now, we can completely

ignore the horizontal details.

-- The bullet is fired with an initial vertical velocity component of 50.0 meters/second.

-- The bullet will continue ascending for 50.0/10 = 5.00 seconds, before it

runs out of gas, surrenders to gravity, and begins to descend.

Additional factoids that emerge from the work but are not required for full credit

on the exam question:

-- The gun is aimed 30 degrees above the horizontal, and the bullet leaves it at 100 meters/second.

-- The highest altitude above ground achieved by the bullet is (125 meters) + (altitude of the gun muzzle).

-- Horizontal distance the bullet travels at or above the altitude of the gun muzzle is

(86.6 m/s) x (10 sec) = 866 meters. -- The Earth's curvature, and the effects of air resistance, are ignored to make

the calculation easier.

-- We're dealing with a very slow bullet here. Even the mild-mannered 22-caliber

"Short round" , with a 1.9-gram bullet and either a small amount of gunpowder

or none at all, intended only for indoor training or target practice, has a typical

muzzle velocity of around 210 meters/second ... double that of the fearsome

projectile described in the question.

Answer 2

The rest of the question wasn't provided. The amount of time it would take before the bullet gets to the highest point of its trajectory would by 5 s.