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Today, 02:02

2. A cannonball is fired at an angle 0 = 35° to the horizontal, with a velocity

of 85 ms 1.

a) What is the maximum vertical height reached by the cannonball?

b) What is the maximum horizontal range of the cannonball?

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Answers (1)
  1. Today, 02:09
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    a) 120 m

    b) 690 m

    Explanation:

    Given:

    x₀ = 0 m

    y₀ = 0 m

    v₀ = 85 m/s

    θ = 35°

    aₓ = 0 m/s²

    aᵧ = - 9.8 m/s²

    a) At the maximum height, the vertical velocity vᵧ = 0 m/s. Find y.

    vᵧ² = v₀ᵧ² + 2aᵧ (y - y₀)

    (0 m/s) ² = (85 sin 35° m/s) ² + 2 (-9.8 m/s²) (y - 0 m)

    y ≈ 120 m

    b) At the maximum range, the cannonball lands, so y = 0.

    First, we need to find the time it takes to land.

    y = y₀ + v₀ᵧ t + ½ aᵧ t²

    (0 m) = (0 m) + (85 sin 35° m/s) t + ½ (-9.8 m/s²) t²

    t = 0 s, 9.95 s

    Now find the final horizontal position x:

    x = x₀ + v₀ₓ t + ½ aₓ t²

    x = (0 m) + (85 cos 35° m/s) (9.95 s) + ½ (0 m/s²) (9.95 s) ²

    x ≈ 690 m
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