A stone is thrown upwards from the edge of a 5.3 m high cliff. It just misses the cliff on the way down and hits the ground below with a speed of 25.4 m/s. With what velocity (in m/s) was it released?
To find the velocity at which the stone was released, we can use the principle of conservation of energy. The total mechanical energy of the stone at any point during its motion is the sum of its potential energy and kinetic energy.
At the top of its motion, when it just misses the cliff, the stone has no kinetic energy and only potential energy due to its position above the ground. Therefore, at that point, the total mechanical energy is equal to the potential energy.
The potential energy of an object of mass m at height h above the ground is given by the formula:
PE = m * g * h
where m is the mass of the stone, g is the acceleration due to gravity (approximately 9.8 m/s² on Earth), and h is the height of the cliff (5.3 m in this case).
So, the potential energy of the stone at the top of its motion is:
PE = m * g * h
Now, at the bottom of its motion when it hits the ground, the stone has no potential energy but only kinetic energy due to its speed. The kinetic energy (KE) of an object of mass m moving at velocity v is given by the formula:
KE = (1/2) * m * v²
where v is the velocity of the stone (25.4 m/s in this case).
Since energy is conserved, the potential energy at the top equals the kinetic energy at the bottom:
PE = KE
m * g * h = (1/2) * m * v²
We can cancel out the mass of the stone (m) from both sides:
g * h = (1/2) * v²
Now, we can solve for the initial velocity (v) by rearranging the equation:
v² = 2 * g * h
v = √(2 * g * h)
Substituting the given values:
v = √(2 * 9.8 m/s² * 5.3 m)
v ≈ √(103.88)
v ≈ 10.19 m/s
Therefore, the stone was released with a velocity of approximately 10.19 m/s.