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Mathematics 15 Online
OpenStudy (anonymous):

an archer stands 40.0m from the target. if the arrow is shot horizontally with a velocity of 90.0 m/s, how far above the bull's eye must she aim to compensate for gravity pulling her arrow downward?

OpenStudy (amistre64):

It takes less then .5 secs for it to reach the bulls eye, so my best guess is just aim for the bulls eye :)

OpenStudy (anonymous):

80/81 m. is it correct

OpenStudy (anonymous):

Essentially, we need to determine how far gravity pulls the arrow during the time it takes to hit the target. The arrow is in flight for .444 seconds (40m/90m/s). Use the following equation, which can be derived by differential equations: x-x0 = v0*t+.5a*t^2. Since the arrow was shot horizontally (all velocity goes into the x-direction), we know that there is no initial velocity in the y-direction. Thus, v0 = 0, and the eqn becomes x-x0=.5*a*t^2. Assuming the arrow was fired on earth, a = 9.81m/s^2. Plug this and time into the eqn, and you get x-x0=0.968m. This is the distance the target needs to be placed beneath the arrow. Think 1m is too large? Drop a coin, and measure how far it falls in .45 seconds.

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