When an object is thrown upward, it rises to a certain height before falling back down. Let's denote the initial height as h.
To determine how high the object rises, we can use the equation for vertical motion under constant acceleration. In this case, the object is thrown vertically upward against the force of gravity, so the acceleration is equal to the acceleration due to gravity (approximately -9.8 m/s², taking downward as the negative direction).
The equation that relates the final height (hf), initial velocity (vi), and acceleration (a) is given by:
hf = vi² / (2 * |a|)
Since the object is thrown upward, the initial velocity is positive, so vi = 19.6 m/s (as mentioned in the previous question). Plugging in the values, we have:
hf = (19.6 m/s)² / (2 * 9.8 m/s²) = 19.6² / (2 * 9.8) = 384.16 / 19.6 ≈ 19.6 meters
Therefore, the object rises to a height of approximately 19.6 meters.
Now, let's determine how high the object has risen when it has lost one-third of its original kinetic energy.
The kinetic energy (KE) of an object is given by:
KE = (1/2) * m * v²
Where m is the mass of the object and v is its velocity.
Since the object is thrown vertically upward, it reaches its highest point when its velocity becomes zero. At this point, all the initial kinetic energy is converted into potential energy (due to its position in the Earth's gravitational field).
When the object has lost one-third of its original kinetic energy, it means that two-thirds of the initial kinetic energy has been converted into potential energy.
Therefore, the height at which the object has lost one-third of its original kinetic energy is equal to two-thirds of the maximum height reached.
Hence, the height at which the object has lost one-third of its original kinetic energy is:
(2/3) * 19.6 meters ≈ 13.1 meters