We're given the following:
- The rock's mass (m) is 3.00 kg
- The height (h) at which we observe the rock is 14.3 m
- The observed speed (v) of the rock at that height is 24.7 m/s
- The acceleration due to gravity (g) is 9.8 m/s²
First, we need to calculate the gravitational potential energy (PE) at the height of 14.3m. The formula for calculating gravitational potential energy is:
PE = m*g*h
Substituting the given values, we get:
PE = 3.00 kg * 9.8 m/s² * 14.3 m = 420.42 Joules
Next, we calculate the kinetic energy (KE) when its speed is 24.7 m/s. The formula for kinetic energy is:
KE = 0.5 * m * v²
Substituting the given values, we get:
KE = 0.5 * 3.00 kg * (24.7 m/s)² = 915.135 Joules
Now, according to the work-energy theorem, the work done in throwing the rock up is equal to the change in its kinetic energy. Assuming that the rock was initially at rest, the change in kinetic energy is the same as the sum of the potential energy and kinetic energy at the height of 14.3 m.
KE_initial = PE + KE
Substitute the calculated values:
KE_initial = 420.42 Joules + 915.135 Joules = 1335.555 Joules
So, the initial kinetic energy of the rock just as it left the ground is 1335.555 Joules.
Finally, we can find the initial speed of the rock as it leaves the ground. The formula relating kinetic energy and mass to speed is:
v = √(2 * KE_initial / m)
Substitute the values:
v = √(2 * 1335.555 Joules / 3.00 kg) = 29.84 m/s
So, the rock's speed just as it left the ground is 29.84 m/s.