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

Question

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

Angle = theta = 45 degrees above the horizontal Range = R =D=220 m g =9.8 m/s^2 R =u^2sin 2*theta /g R=u^2sin 2*45 /g R = u^2 /g u = sq rt Rg u = sq rt 220*9.8 = sq rt 2156 u =46.433 m/s Time of flight = T =2usin theta /g T=2u sin45/g =2*46.433 *0.707./9.8=6.699 s Time to fall 6 m = t = sq rt [2h/g] =sq rt [12/9,8] t =1.1065 s Time after the apple be dropped =T - t =6.699 -1.1065 =5.592 s later a) For the arrow to pierce the apple as the arrow hits the tree,the apple be dropped 5.592 s after the arrow was shot

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