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

An arrow is shot at an angle of above the horizontal. The arrow hits a tree at an angle 45 degrees horizontal distance D = 220 away, at the same height above the ground as it was shot. Use g = 9.8 for the magnitude of the acceleration due to gravity.

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 2:02pm
at 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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