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

Question

An arrow is shot at an angle of theta = 45 degree 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 t_a, the time that the arrow spends in the air. Answer numerically in seconds, to two significant figures. 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 angle be dropped, in order for arrow to pierce the apple as the arrow hits the tree? Express your answer numerically in seconds, to two significant figures.

Explanation / Answer

Given that

angle =45 degree

horizontal distance R=220 m

now we find the time of flight

time of flight T=[2Rtan(theta)/g]^1/2

=[2*220*tan45/9.8]^1/2=6.7 sec

now we find the time of desent

time of dasent td=[2H/g]^1/2=[2*6/9.8]^1/2=1.11 se

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