Final answer:
When the pebble is pulled back 40.0 cm against the elastic band, it will go as high as a height of 20.0 m.
Step-by-step explanation:
To solve this problem, we can use the concept of potential energy. When the pebble is pulled back against the elastic band, it gains potential energy. The potential energy can be calculated using the formula PE = 1/2 k x^2, where PE is the potential energy, k is the spring constant, and x is the displacement of the pebble from its equilibrium position.
In this case, when the pebble is pulled back 20.0 cm (or 0.2 m), it goes 6.0 m high. We can use this information to find the spring constant k. Rearranging the formula, we have k = 2 * PE / x^2. Substituting the values, we get k = 2 * 50 / (0.2)^2 = 5000 N/m.
Now, if we pull the pebble back 40.0 cm (or 0.4 m), we can calculate the potential energy at that position using the same formula and the newly found spring constant. Rearranging the formula, we have PE = 1/2 k x^2. Substituting the values, we get PE = 1/2 * 5000 * (0.4)^2 = 200 J.
Since the pebble gains potential energy when pulled back, it will reach a higher height when launched. Using the principle of conservation of energy, we can say that the potential energy gained is equal to the potential energy at the highest point. Therefore, when launched with a potential energy of 200 J, the pebble will go as high as a height of 20.0 m.