A projectile is fired at 0=433.0 m/s at an angle of =64.8∘, with respect to the horizontal. Assume that air friction will shorten the range by 28.1%. How far will the projectile travel in the horizontal direction, ?

Here, we are required to determine how far the projectile will travel in the horizontal direction if it's projected according to the parameters in the question above.

The distance travelled by the projectile in the horizontal direction is,

R' = 10598.8m.

Explanation:

How far a projectile travels in the horizontal direction is termed it's RANGE.

The formula for calculating the range is, R = (u²Sin2θ)/g

Where u = velocity/speed of projection,

θ = angle of projection with respect to the horizontal,

And, g = acceleration due to gravity (9.8ms⁻²).

Therefore, the range, R in this case is

R = {433² × Sin (2×64.8°)}/ 9.8

Therefore, the range of the projectile under ideal circumstances of 'no air friction' would be,

R = 14,741m.

However, air friction shortens the range by 28.1%.

Therefore, the actual distance travelled in the horizontal direction is,

R' = {(100-28.1) × 14,741} / 100

Therefore, the distance travelled by the projectile in the horizontal direction is,

R' = 10598.8m.

Read more:

link

10,600 m

Explanation:

Use range equation:

R = v₀² sin(2θ) / g

R = (433.0 m/s)² sin(2 × 64.8°) / (9.8 m/s²)

R = 14,700 m

The range is shortened by 28.1%.

0.719R = 10,600 m

area of focus