A 74.1 kg high jumper leaves the ground with a vertical velocity of 9.7 m/s.
How high can he jump? The acceleration
of gravity is 9.8 m/s
2
.
Answer in units of m

4.80 m

Explanation:

We are given the mass of the high jumper, its initial velocity, and the acceleration of gravity. We are trying to find the vertical displacement of the high jumper.

Let's set the upwards direction to be positive and the downwards direction to be negative.

List out the relevant known variables.

We still need one more variable in order to use the constant acceleration equations. Since we are trying to find the max height of the jumper, we can use the fact that at the top of its trajectory, its final velocity will be 0 m/s.

    4. v = 0 m/s

Using these four variables, let's find the constant acceleration equation that contains these variables:

Substitute the known values into the equation and solve for Δx.

The high jumper can jump to a max height of 4.80 m.

frequency or wavelenth of em radiation in a vacuum

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.