G A 0.56-kg ball is thrown vertically upward with an initial speed of 17.0 m/s. What is the potential energy of the ball when it has travelled one-half of the distance to its maximum height? (Use the launch point to be the zero reference point of measurements of the heights)

Explanation:

Given that

Mass of ball=0.56kg

Initial speed=17m/s

Let g=9.81m/s²

Potential energy when the ball travel half way=?

Using equation of free fall to know the maximum height the ball will reach

v²=u²-2gH.

Upward direction i.e against gravity

The the velocity at the maximum height is zero because the body will momentary stop before coming downward , therefore v=0

Therefore,

v²=u²-2gH

0²=17²-2×9.81×H

0=289-19.62H

Then, 19.62H=289

H=289/19.62.

H=14.73m

Then,

The half way is distance is ½×H=½×14.73=7.36m

Then the potential energy at this point is given as

P.E=mgh

P.E=0.56×9.81×7.36

P.E=40.43J

The potential energy halfway is 40.43J

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).