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An arrow is shot at an angle of ?=45? 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/s2 for the magnitude of the acceleration

due to gravity.

a. Find ta, 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?

Express your answer numerically in seconds, to two significant

figures.

Edit
Added Wed, 06 Jul '16
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a)

The arrow drops to the same horizontal plane from which it was

shot.

R = u T cos ? and T = 2 u sin ?

R tan ? = gt� / 2

220 tan 45 =4.9 t�

t = 6.7 s

===========================

b)

Assuming that this is the continuation of the previous problem and

not a separate problem,

The arrow takes 6.7 s to hit the tree at the same horizontal

plane.

The time taken by the apple to fall through a height of 6 m is

given by

t = ? (2h/g) = ? (12/ 9.8) = 1.11 s.

Difference in time is 6.7 � 1.11 =5.59 s

Hence the apple should be dropped from a height of 6 m from the

point where the arrow hits the tree after a time of 5.59 s the

arrow was shot.

Edit
Added Wed, 06 Jul '16
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a) d = (v^2/g)sin(2*theta)

here d =220m

g =9.81 m/s^2

theta = 450

so v= [(220*9.81)/sin(2*theta)]1/2

v= 46.456 m/s

so for horizontal distance covered

vcos(theta)*t = d

or t = d/vcos(theta)

or t = 220/(46.456*{1/2}1/2)

or t= 6.697 secs

b) time taken by apple to cover 6meters will be

s = ut 1/2gt^2

u=0

so t = (2*s/g)1/2

so t = 1.106 secs

so apple should be dropped after (6.697-1.106)secs

or after 5.591 secs

Edit
Added Wed, 06 Jul '16
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