Final answer:
The initial velocity of the rock thrown vertically upwards can be calculated using the conservation of mechanical energy. By equating the sum of kinetic and potential energy at 16.0 meters to the initial kinetic energy at the ground and solving for the initial velocity, one can determine the speed at which the rock was initially thrown.
Step-by-step explanation:
To calculate the initial velocity of the rock when it was thrown into the air, we use the principles of kinematics and energy conservation. Since the force of gravity is the only force doing work on the rock (assuming negligible air resistance), the rock's mechanical energy is conserved. The two main forms of mechanical energy to consider here are gravitational potential energy and kinetic energy.
Gravitational potential energy (U) at height h is given by U = mgh, where m is mass, g is the acceleration due to gravity (9.8 m/s²), and h is the height above the reference point (ground level in this case). Kinetic energy (KE) is given by KE = ½mv², where v is the velocity of the rock.
At 16.0 meters above the ground, the rock has both gravitational potential energy and kinetic energy due to its velocity of 23.0 m/s. Therefore, we can write that the sum of potential and kinetic energy at this point must be equal to the initial kinetic energy when the rock was thrown from the ground, as no other energy has been added or lost (energy conservation). Using the formula for energy conservation and solving for the initial velocity, we find:
½mv²_initial + mgh = ½mv² + mgh_at_16m
Since the mass m cancels out, we simplify to:
½v²_initial = ½(23.0m/s)² + (9.8m/s²)(16.0m)
Calculating the right side and then multiplying by 2 gives us the initial velocity squared. Taking the square root will provide us with the initial velocity.