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Today, 01:21

A projectile is launched at ground level with an initial speed of 53.0 m/s at an angle of 35.0° above the horizontal. It strikes a target above the ground 2.50 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

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  1. Today, 03:26
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    x = 108.5 m y = 45.37 m,

    Explanation:

    This is a projectile launching exercise, the horizontal distance it reaches is called the range and can be found by the expression

    x = v₀ₓ t

    the initial hairiness is

    v₀ₓ = v₀ cos θ

    x = v₀ cos θ t

    let's calculate

    x = 53 cos 35 2.50

    x = 108.5 m

    the height of the projectile can be calculated with

    y = v₀ t - ½ g t²

    y = vo sin tea t - ½ g t²

    y = 53 sin 35 2.50 - ½ 9.8 2.5²

    y = 45.37 m
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