A ball is thrown straight up and rises to a maximum height of 24 m above the ground. At what height is the speed of the ball equal to half of its initial value? Assume that the ball starts at a height of 1.9 m above the ground.

18.615

To determine the height at which the speed of the ball is equal to half of its initial value, we can use the principle of conservation of mechanical energy.

The initial mechanical energy of the ball is equal to the sum of its potential energy and kinetic energy. At its maximum height, all of the ball's initial kinetic energy is converted into potential energy, so we can equate the two:

Initial KE = Final PE

The initial kinetic energy (KE) is given by:

KE = 0.5 * mass * initial velocity^2

Since the ball starts at a height of 1.9 m above the ground, the initial potential energy (PE) is given by:

PE = mass * g * initial height

Where mass is the mass of the ball and g is the acceleration due to gravity.

At the height where the speed of the ball is equal to half of its initial value, the kinetic energy is reduced by half. Let's denote this height as h. At this point, the kinetic energy of the ball is given by:

KE' = 0.5 * mass * (0.5 * initial velocity)^2 = 0.125 * mass * initial velocity^2

The potential energy at this height is given by:

PE' = mass * g * (initial height + h)

Now, we can equate the initial kinetic energy to the final potential energy at height h:

0.5 * mass * initial velocity^2 = mass * g * (initial height + h)

We can simplify the equation by canceling out the mass:

0.5 * initial velocity^2 = g * (initial height + h)

To solve for h, we isolate it on one side of the equation:

h = (0.5 * initial velocity^2 / g) - initial height

Now, we have all the necessary values to calculate h. Substitute the given values for initial velocity, g, and initial height, and solve for h:

h = (0.5 * initial velocity^2 / g) - initial height

Note: Make sure to convert the units to be consistent throughout the calculations.