Question

A bow has k= 458 N/m. It is

pulled back 0.720 m, and used tofire a 0.0925 kg arrow. It hits a bird 24.7 m above the ground. What is the speed of the arrow when it hits?

Answer 1

The final speed of the arrow when it hits the target is 45.63 m/s.

Step-by-step 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.

`(1)/(2) kx^2 = (1)/(2) mu^2\\\\kx^2 = mu^2\\\\u^2 = (kx^2)/(m) \\\\u= \sqrt{(kx^2)/(m) } \\\\u = 50.66 \ m/s`

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.