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