Work and Energy Question

Question

 An archery bow is drawn a distance d = 0.49 m and loaded with an arrow of mass m = 0.071 kg. The bow acts as a spring with a spring constant of k = 167 N/m, and the arrow flies with negligible air resistance. To simplify your work, let the gravitational potential energy be zero at the initial height of the arrow.

if the arrow is shot at an angle of θ = 45° above the horizontal, how high, in meters above the initial height, will the arrow be when it reaches its peak? 

Answer 1

elastic potential energy turns into the kinetic energy of the arrow

1/2 k x^2 = 1/2 m v^2

167 * 0.49^2 / 0.071 = v^2

v = 23.8

This will be the initial velocity for the projectile motion.

At the highest point, V_y = 0. Initial V_y = 23.8 * sin 45 = 16.8

V^2 - V_i^2 = 2 a d

0 - 16.8^2 = 2 * (-9.8) h

h= 14.4 m