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

Question

An arrow is shot at an angle of heta = 45degrees 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.

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


b.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

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

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