An archer shoots an arrow with a velocity of 45.0m/s at an angle of 50.0degrees with the horizontal.An assistant standing on the level ground 150m downrange from the launch point throws an apple straight up with the minimum intial speed necessary to meet the path of the arrow.(a) what is the initial speed of the apple? (b) at what time after the arrow launch should the apple be thrown so that the arrow hits the apple?

Answer 1

Vo = 45m/s @ 50o
Xo = 45*cos50 = 28.93 m/s.
Yo = 45*sin50 = 34.47 m/s.

a. Y^2 = Yo^2 + 2g*h.
h = (Y^2-Yo^2)/2g.
h = (0-(34.47)^2)/-19.6=60.6 m.=Max ht.

Y = Yo + g*t.
Tr = (Y-Yo)/g = (0-34.47)/-9.8=3.52 s.=
Rise time or time to reach max ht.

Xo*T = 150m
28.93*T = 150
T = 5.18 s. = Time in air.

Tf = 5.18 - 3.52 = 1.66 s.

h = hmax - 0.5g*t^2
h = 60.6 - 4.9*(1.66)^2=60.6-13.5=47.1
m above launching level.

d = Vo*t + 0.5g*t^2 = 47.1 m.
Vo*5.18 - 4.9*(5.18)^2 = 47.1
Vo*5.18 - 131.48 = 47.1
Vo*5.18 = 178.58
Vo = 34.47 m/s. = Initial velocity of the apple.

b. T = 0. Apple must be thrown at same time arrow is launched.

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Answer 2

An arrow is shot from an archer’s bow at an angle of 45° with a velocity of 30 m/s. It hits a bullseye on a target after 2 seconds. At what vertical and horizontal displacement is the arrow after that time?

Answer 3

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