1. An archer shoots an arrow horizontally at a target 12 m away. The arrow is aimed directly at the
center of the target but

1. An archer shoots an arrow horizontally at a target 12 m away. The arrow is aimed directly at the
center of the target but it hits 20 cm lower. What is the initial speed of the arrow?
Given:
Required:
Solution:
Final Answer:​

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  1. Answer:

    I used d=v0t+.5(-9.8)t^2, with-.056=.5(-9.8)t^2 to find t, which I would plug into the horizontal part of the equation.

    Then, I took the square root of both sides and divided sqrt.056 by .5(9.8). This resulted in .011

    d = (1/2)gt2 = 0.56 m

    Get t from this, and then set

    v0t = 15 m [i.e. the horizontal travel distance at constant speed],

    and then get v0 knowing t.

    (1/2)(9.8)t2 = 0.56

    t = √[2(0.56)/9.8] s ≅ 0.338 s

    v0t = 15 m

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