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5 August, 16:17

A ball is thrown so that its initial vertical and horizontal components of velocity are 30 m/s and 15 m/s, respectively. Estimate the maximum height the ball reaches. (Use 10 m/s2 as the acceleration of gravity.) 101 Incorrect: Your answer is incorrect.

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  1. 5 August, 18:45
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    H = 45 m

    Explanation:

    First we find the launch velocity of the ball by using the following formula:

    v₀ = √ (v₀ₓ² + v₀y²)

    where,

    v₀ = launching velocity = ?

    v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s

    v₀y = Vertical Component of Launch Velocity = 30 m/s

    Therefore,

    v₀ = √[ (15 m/s) ² + (30 m/s) ²]

    v₀ = 33.54 m/s

    Now, we find the launch angle of the ball by using the following formula:

    θ = tan⁻¹ (v₀y/v₀ₓ)

    θ = tan⁻¹ (30/15)

    θ = tan⁻¹ (2)

    θ = 63.43°

    Now, the maximum height attained by the ball is given by the formula:

    H = (v₀² Sin² θ) / 2g

    H = (33.54 m/s) ² (Sin² 63.43°) / 2 (10 m/s²)

    H = 45 m
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