Abasketball rolls onto the court with a speed of 4 . 0 m s 4.0 s m 4, point, 0, space, start fraction, m, divided by, s, end fraction to the right, and slows down with a constant acceleration of 0 . 5 0 m s 2 0.50 s 2 m 0, point, 50, space, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction over 1 4 m 14m14, space, m. what is the velocity of the basketball after rolling for 1 4 m 14m14, space, m?

1.4 m/s

Explanation:

The final velocity of the ball can be found by using the following SUVAT equation:

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance travelled

In this problem,

u = 4.0 m/s

d = 14 m

a = -0.50 m/s^2

Solving the equation for v,

come closer and become far apart and become steam/gas

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.