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 maxium height is 34m/s