Explaination
the maximum hight is reached by the arrow is 4.40m
From Newton's second law and third equation of motion, the maximum speed of the arrow at the instant it leaves the bow is 11.8 m/s and the maximum height reached by the arrow in its flight into the air is 7.08 m
According to Newton's second law: The rate of change in momentum is directly proportional to the force applied. That is,
F = ma
Given that an archer uses an average force of 50.0 N to draw the string of his bow through a distance of 0.412 m. Then, he fires a 297 g arrow straight up into the air.
The parameters given are;
The maximum speed of the arrow at the instant it leaves the bow can be calculated by first calculating the acceleration.
F = ma
Substitute all the necessary parameters
50 = 0.297a
a = 50 / 0.297
a = 168.4 m/[tex]s^{2}[/tex]
by using third equation of motion
[tex]V^{2}[/tex] = [tex]U^{2}[/tex] + 2as
Since it is starting from rest, U = 0
[tex]V^{2}[/tex] = 2 x 168.4 x 0.412
[tex]V^{2}[/tex] = 138.72
V = [tex]\sqrt{138.72}[/tex]
V = 11.8 m/s
The maximum height reached by the arrow in its flight into the air will be calculated by the same formula where acceleration a = g
[tex]V^{2}[/tex] = [tex]U^{2}[/tex] - 2gH
At maximum height, final velocity V = 0
0 = [tex]11.8^{2}[/tex] - 2 x 9.8 H
138.7 = 19.6H
H = 138.7 / 19.6
H = 7.08 m
Therefore, the maximum speed of the arrow at the instant it leaves the bow is 11.8 m/s and the maximum height reached by the arrow in its flight into the air is 7.08 m
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