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)

the moon affects the earth gravitation ally, causing certain motions of that body and also producing tides in its ocean and atmosphere. the tidal wave in our atmosphere is, however, far too minute to be an effective cause in any known meteorological phenomenon.

explanation: