• a rocket is launched straight up from the earth's surface at a speed of 1.90×104 m/s .

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  • How high does it go? Assuming no air resistance or other outside forces, the rocket would reach a height of 4.19 × 108 m after 459.72 seconds.

    • Answered:

      Raymond May

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  • Assuming no air resistance and a constant downward acceleration of 9.81 m/s2 due to gravity, the time it takes for the rocket to reach its maximum height can be calculated using the formula s = u*t + 1/2 * a * t2, where s is the displacement, u is the initial velocity, and a is the acceleration. Plugging in the values from the given problem, we get s = 1.90×104 m/s * t + 1/2 * 9.81 m/s2 * t2. The maximum height is found when the rocket's velocity is 0, so we set this equation equal to 0, and solve for t: 0 = 1.90×104 m/s * t + 1/2 * 9.81 m/s2 * t2. We can rearrange this equation to t2 + 20.51t – 1.90×104 = 0, and solve for t by using the quadratic formula. The solution is t = 20.50 s. This is the time it takes for the rocket to reach its maximum height.

    • Answered:

      Weston Baldwin

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