A cannonball is fired on flat ground at 420 m/s at a 53.0° angle. how far away does it land?

17,300 m

Explanation:

Using kinematic equations, first find the time it takes to land.

Δy = v₀ t + ½ at²

0 m = (420 sin 53.0° m/s) t + ½ (-9.8 m/s²) t²

t = 0 s or 68.5 s

The horizontal distance it moves in that time is:

Δx = v₀ t + ½ at²

Δx = (420 cos 53.0° m/s) (68.5 s) + ½ (0 m/s²) (68.5 s)²

Δx = 17,300 m

Alternatively, you can use the range equation:

R = v₀² sin(2θ) / g

R = (420 m/s)² sin(2 × 53.0°) / (9.8 m/s²)

R = 17,300 m

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

im srry but ur wrong the answer is d a smooth rock, some red dirt, and an empty bird's nest.