Answer:
The final speed of the arrow when it hits the target is 45.63 m/s.
Explanation:
Given;
spring constant of the bow, k = 458 N/m
extension of the bow, x = 0.72 m
mass of the arrow, m = 0.0925 kg
let the initial speed of the arrow after being fired = u
Apply the law of conservation of energy, the elastic potential energy of the bow will be converted to kinetic energy of the arrow.
[tex]\frac{1}{2} kx^2 = \frac{1}{2} mu^2\\\\kx^2 = mu^2\\\\u^2 = \frac{kx^2}{m} \\\\u= \sqrt{\frac{kx^2}{m} } \\\\u = 50.66 \ m/s[/tex]
The speed of the arrow when it hits a target 24.7 m above the ground is calculated as;
v² = u² + 2gs
where;
v is the final speed when the arrow hits the target
g is acceleration due to gravity = (-9.8 m/s², upward motion)
v² = 50.66² + 2(-9.8)24.7
v² = 2566.44 - 484.12
v² = 2082.32
v = √2082.32
v = 45.63 m/s
Therefore, the final speed of the arrow when it hits the target is 45.63 m/s.
