A 12.0-g ball is placed in a slingshot with a spring constant of 200 N/m. The ball is stretched back 0.500 m. The slingshot is aimed straight up, then fired. What is the maximum height that the ball will reach?

213 m

106 m

98 m

182 m

The energy put into the slingshot will be translated to the ball. This energy is

(1/2) kx² = (1/2) (200N/m) (0.5) ² = 25J

This will be converted entirely to gravitational potential energy at the peak of the ball's height. This is given by mgh. So mgh = 25J. h=25J / (0.012kg·9.81m/s²) = 212.37. So 213m is the best answer