An arrow is shot at an angle of =45degree above the horizontal. The arrow hits a

Question

An arrow is shot at an angle of =45degree above the horizontal. The arrow hits a tree ahorizontal distance D = 220 m away, at the same heightabove the ground as it was shot. Use g = 9.8m/s^2 for themagnitude of the acceleration due to gravity. The time that thearrow spends in the air is 6.7s. Suppose someone drops an apple from avertical distance of 6.0 meters, directly above the point where thearrow hits the tree. How long after the arrow was shot should theapple be dropped, in order for the arrow to pierce the apple as thearrow hits the tree? An arrow is shot at an angle of =45degree above the horizontal. The arrow hits a tree ahorizontal distance D = 220 m away, at the same heightabove the ground as it was shot. Use g = 9.8m/s^2 for themagnitude of the acceleration due to gravity. The time that thearrow spends in the air is 6.7s. Suppose someone drops an apple from avertical distance of 6.0 meters, directly above the point where thearrow hits the tree. Suppose someone drops an apple from avertical distance of 6.0 meters, directly above the point where thearrow hits the tree. How long after the arrow was shot should theapple be dropped, in order for the arrow to pierce the apple as thearrow hits the tree? How long after the arrow was shot should theapple be dropped, in order for the arrow to pierce the apple as thearrow hits the tree?

Explanation / Answer

Range = horizontal velocity * time of flight         D =v2sin2 / g Fom this we get initial velocity         v = (Dg /sin2) =   (220*9.8 / sin90) = 46.43m/s Time of ascent ta = v sin / g                            = 46.43*0.707 / 9.8 = 3.35 s (b) Total time of flight T = 2ta = 6.70s

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