A rocket 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)
1 answer:
Answer:
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)
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