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2 August, 15:38

An arrow is shot at an angle 38 ◦ with the horizontal. It has a velocity of 46 m/s. How high will the arrow go?

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  1. 2 August, 16:03
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    40·919 m

    Explanation:

    Initial velocity of the arrow = 46 m/s

    Angle at which it is thrown from horizontal = 38°

    At the maximum height, the vertical component of velocity will be 0

    Initial velocity in vertical direction = 46 * sin (38) = 28·32 m/s

    From the formula

    v² - u² = 2 * a * s

    where

    v is the final velocity

    u is the initial velocity

    a is the acceleration

    s is the displacement

    Considering the formula in vertical direction and taking upward direction as positive

    v = 0

    u = 28·32 m/s

    a = - g = - 9·8 m/s²

    Let s be the maximum height

    - 28·32² = - 2 * 9.8 * s

    ⇒ s = 40·919 m

    ∴ The arrow will go 40·919 m high
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