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August 31, 2014

August 31, 2014

Posted by **Nami** on Friday, September 28, 2012 at 1:19pm.

Find t_a, the time that the arrow spends in the air.

Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree.

How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree?

Please provide formulas and numeric answers

Thank You!

- Physics -
**bobpursley**, Friday, September 28, 2012 at 2:02pmat 45degrees, the final vertical velocity is the same as the horizontal velocity. So the same applies to launch.

Horizontal analysis:

distance=Vhorizontal*timeinair.

220=Vhorizontal*time

time=220/Vh

Vertical analysis:

hf=hi+Vv*t-1/2 g t^2

0=0+Vv*220/Vh-4.9 (220/Vh)^2

because Vv=Vh

220=4.9*220^2/Vh^2

solve for Vh

then, solve for timeinair from the first equation.

How long does it take for the apple to fall? 6=gt or t=6/g

subtract that from the time in air above, and you have the time after launch for the apple to be released.

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