- #1

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## Homework Statement

Someone ask me a problem found in college physics textbook. It states: An arrow is shot at an angle of 45 degree above the horizontal. The arrow hits a tree a horizontal distance away D=220m, at the same height above the ground as it was shot. Use g for the magnitude of the acceleration due to gravity. Find the time with the arrow is traveling in the air.

**2. The attempt at a solution**

Assum the total time when the arrow travel in the air is T. First, the let the initial velocity be V and the initial horizontal velocity is [tex]V_{x0}[/tex] and vertical velocity is [tex]V_{y0}[/tex]. We have

[tex]V_{x0}=V_{y0} = D/T[/tex]

For y position (height), when the arrow hit the tree, we have

[tex]D = V_{y0}T - gT^2/2[/tex]

But [tex]V_{y0}=V_{x0}[/tex], this gives [tex]gT^2/2=0[/tex] ?

I don't know what's going on here. So I assume the vertical velocity when the arrow hit the tree is ZERO, so

[tex]0 = V_{y0} - gT = V_{x0} - gT = D/T - gT[/tex]

which gives [tex]T=\sqrt{D/g}[/tex]

I know the correct answer should be [tex]T=\sqrt{2D/g}[/tex],I just don't know what's going on here. I check many times, please tell me where I get the problem wrong.