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14 April, 19:10

A cannonball is fired on flat ground

at 420 m/s at a 53.0° angle. What

maximum height does it reach?

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Answers (1)
  1. 14 April, 20:34
    0
    hmax = 5740.48 m. The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.

    This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.

    V₀ = 420m/s and θ₀ = 53.0°

    So, when the cannonball is fired it has horizontal and vertical components:

    V₀ₓ = V₀ cos θ₀ = (420m/s) (cos 53°) = 252.76 m/s

    V₀y = V₀ cos θ₀ = (420m/s) (cos 53°) = 335.43m/s

    When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:

    Vy = V₀y - g tₐ = 0

    tₐ = V₀y/g

    tₐ = (335.43m/s) / (9.8m/s²) = 34.23s

    Then, the maximum height is reached in the instant tₐ = 34.23s:

    h = V₀y tₐ - 1/2g tₐ²

    hmax = (335.43m/s) (34.23s) - 1/2 (9.8m/s²) (34.23s) ²

    hmax = 11481.77m - 5741.29m

    hmax = 5740.48m
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