## College Physics (4th Edition)

The pebble flies up to a height of $12.8~meters$ above the starting point.
By conservation of energy, the gravitational potential energy at maximum height will be equal to the elastic potential energy stored in the catapult initially: $mgh = \frac{1}{2}kx^2$ $h = \frac{kx^2}{2mg}$ $h = \frac{(320~N/m)(0.20~m)^2}{(2)(0.051~kg)(9.80~m/s^2)}$ $h = 12.8~m$ The pebble flies up to a height of $12.8~meters$ above the starting point.