Arocket is fired at an initial speed v0 = 151.0 m/s from ground level, at an angle θ = 46 degrees above the horizontal. a wall is located at d = 41.0 m. its heigh is h = 30.0 m. ignore air resistance. the magnitude of the gravitational acceleration is 9.8 m/s2. choose the right as positive x-direction. choose upward as psotitive y-direction keep 2 decimal places in all answers find v0x, the x component of the initial velocity (in m/s)

v0x = 104.89 m/s

The vector v0x will be:

v0x = (104.89 m/s, 0)

Explanation:

Please, see the attached figure for a better understanding of the problem.

To obtain v0x, we have to use this trigonometric rule:

cos θ = adjacent / hypotenuse

Seeing the figure, notice that the side adjacent to the angle θ is the x-component of the initial velocity v0. The hypotenuse is the magnitude of the vector v0. Then:

cos θ = magnitude v0x / magnitude v0

magnitude v0 * cos θ = magnitude v0x

151.0 m/s * cos 46 = magnitude v0x

magnitude v0x = 104.89 m/s

The vector v0x will be:

v0x = (104.89 m/s, 0)

a.) is the correct answer.

explanation:

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.