Question

A bullet is fired into the air at an angle of 45°. How far does it travel before it is 1,000 feet above the ground? (Assume that the bullet travels in a straight line; neglect the forces of gravity, and give your answer to the nearest foot.)

Answer 1

The bullet will travel zero distance before reaching a height of 1,000 feet above the ground.

Step-by-step explanation:

To find the horizontal distance traveled by the bullet before it is 1,000 feet above the ground, we need to calculate the range of the projectile. The range of a projectile launched at an angle of 45° can be calculated using the equation:

Range = (initial velocity^2 * sin(2θ))/g

where θ is the angle of the projectile, g is the acceleration due to gravity, and initial velocity is the speed of the projectile. In this case, since we are neglecting the forces of gravity, g can be considered as zero. Therefore, the range simplifies to:

Range = initial velocity^2 * sin(2θ) / 0

Simplifying further, we find:

Range = 0

Accordingly, the bullet will travel zero distance before it reaches a height of 1,000 feet above the ground.

Answer 2

It travels 1414 feets.

Step-by-step explanation:

Let's take the length the bullet travels l as the hypotenuse of a right triangle and the height it reaches one of its sides. Since we got the angle α at which it was fired and the height h it reached, we can calculate l using the sin(α) function:

`sin(\alpha )=(opposite side)/(hypotenuse)\\sin(\alpha)=(h)/(l)\\l=(h)/(sin(\alpha))`

Replacing:

`l=(1000ft)/(sin((\pi)/(4)))`

Solving and roundin to the nearest foot:

`l=1414 ft`