You shoot a \(51-\mathrm{g}\) pebble straight up with a catapult whose spring constant is \(320 \mathrm{N} / \mathrm{m} .\) The catapult is initially stretched by \(0.20 \mathrm{m}\). How high above the starting point does the pebble fly? Ignore air resistance.

We will first find the initial potential energy (\(U_{spring}\)) stored in the spring when it's stretched by the given distance. The potential energy of a spring is given by: $$ U_{spring} = \frac{1}{2}kx^2, $$ where \(k\) is the spring constant, and \(x\) is the stretch in the spring. Given, \(k = 320\,\mathrm{N/m}\) and \(x = 0.20\,\mathrm{m}\). Let's plug in the values and calculate the initial potential energy. $$ U_{spring} = \frac{1}{2} (320\,\mathrm{N/m})(0.20\,\mathrm{m})^2$$

As the pebble is launched, the spring loses its potential energy as it returns to its equilibrium position. The lost potential energy is converted into kinetic energy, which propels the pebble upward. As the pebble rises, it loses kinetic energy and gains gravitational potential energy. At the maximum height, the pebble's velocity reaches zero, meaning all the kinetic energy has been converted into gravitational potential energy. We know that the total mechanical energy of the system is conserved. So, the potential energy stored in the spring in the beginning will be equal to the gravitational potential energy of the pebble at its maximum height: $$ U_{spring} = U_{gravitational} $$ The gravitational potential energy is given by the formula: $$ U_{gravitational} = mgh $$ where \(m\) is the mass of the pebble, \(g\) is the acceleration due to gravity (\(9.8\,\mathrm{m/s^2}\)), and \(h\) is the height of the pebble above its starting point.

Now, we can set up an equation to find the maximum height, \(h\), reached by the pebble: $$ \frac{1}{2}kx^2 = mgh $$ Now plug in the known values: \(k = 320\,\mathrm{N/m}\), \(x = 0.20\,\mathrm{m}\), \(m = 51\,\mathrm{g} = 0.051\,\mathrm{kg}\), and \(g = 9.8\,\mathrm{m/s^2}\). $$ \frac{1}{2}(320\,\mathrm{N/m})(0.20\,\mathrm{m})^2 = (0.051\,\mathrm{kg})(9.8\,\mathrm{m/s^2})h $$ Now solve for \(h\): $$ h = \frac{\frac{1}{2}(320\,\mathrm{N/m})(0.20\,\mathrm{m})^2}{(0.051\,\mathrm{kg})(9.8\,\mathrm{m/s^2})} $$ Calculate the value of \(h\): $$ h \approx 1.27\,\mathrm{m} $$

The pebble reaches a maximum height of approximately \(1.27\,\mathrm{m}\) above its starting point.