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?

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.

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.

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2