An arrow is shot at an angle of theta = 45 degrees above the horizontal. The arr

Question

An arrow is shot at an angle of theta = 45 degrees above the horizontal. The arrow hits a tree a horizontal distance D = 220 m away, at the same height above the ground as it was shot. Use g = 9.8 m/s^2} for the magnitude of the acceleration due to gravity.

Find 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?

Explanation / Answer

QUESTION ONE

Given initial data:

R = 220 m
g = 9.8 m/s^2
? = 45°

Find the initial total velocity:

Vo = SQRT { [R * g] / [sin2?] }
Vo = SQRT { [ (220 m) * (9.8 m/s^2) ] / [sin(2 * 45)] }
Vo = SQRT { [ 2,156 m^2/s^2 ] / [sin90] }
Vo = SQRT { [ 2,156 m^2/s^2 ] / [ 1 ] }
Vo = SQRT { 2,156 m^2/s^2 }
Vo = 46.4 m/s

Find the initial component of horizontal velocity

Vox = Vo * cos?
Vox = (46.4 m/s) * (cos45)
Vox = (46.4 m/s) * (0.707)
Vox = 32.8 m/s

Find the total travel time

Tt = R / Vox
Tt = (220 m) / (32.8 m/s)
Tt = 6.70 s

QUESTION TWO

Given new data:

R = 220 m
g = 9.8 m/s^2
? = 45°
Vo = 46.4 m/s
Vox = 32.8 m/s
Tt = 6.70 s
Y = 6.0 m

Find how far it takes the apple to drop 6 m

t = SQRT { [2Y] / g }
t = SQRT { [2 * (6.0 m)] / (9.8 m/s^2) }
t = SQRT { [ 12 m ] / (9.8 m/s^2) }
t = SQRT { 1.22 s^2 }
t = 1.12 s

Subtract apple fall time from arrow shot time to get time to drop the apple

Ta = Tt - t
Ta = (6.70 s) - (1.12 s)
Ta = 5.58 s

Answer -->

(1) Total arrow travel time = 6.70 s
(2) Drop the apple at time = 5.58 s

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