Physics Question

Question

My question is "An archer shoots an arrow horizontally at a target 15 m away. The arrow is aimed directly at the center of the target, but it hits 56 cm lower. What was the initial speed of the arrow?"

 

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, which is illogical and incorrect, obviously. However, I can't see what I'm doing wrong here. Could someone help me please?

 

Thanks in advance.

Answer 1

Your mistake was to use v₀ in the calculation of height.  The problem statement says the arrow was shot horizontally, so the initial speed in the vertical direction is zero, while the initial speed in the horizontal direction is v₀.  The vertical drop is 

 

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

 

v₀ = (15 m)/(0.338 s) ≅ 44.4 m/s in the horizontal direction, with no initial vertical velocity