A 0.60-kg basketball is dropped out of a window that is 6.1mabove the ground. The ball is caught by a person whose handsare 1.5m above the gound.1. How much work is done on e the ball by its weight?What is the gravitational potential energy of the basketball,relative to the ground, when it is ... 2. released and 3. caught?4. How is the change (PEf- PE0) in the ball'sgravitational potential energy related to the workdone by its weight?

The energy that an object has or acquires when its position changes as a result of being in a gravitational field is known as gravitational potential energy.
Gravitational potential energy can be defined as an energy that has a connection to gravitational force or gravity.

The work that gravity does;

W = F.S = mgΔh

In light of the fact that gravity points downward and that displacement is also downward, their angle is 0 and their cosine 1, respectively.

W = mgΔh

g = 9.8 m/s²

Δh = 4.6 m

m = 0.60 kg

W = 0.60 x 9.80 x 4.6

W = 2.7 J

The source of gravitational potential energy in relation to the ground is;

U = mgh

The release point therefore

Ug = mgh(0)

U0 = 0.60 x 9.8 x 6.1

U0 = 35.87 J

At the point of release

U(gf) = mgh(f)

U(gf) = 0.60 x 9.80 x 1.5

U(gf) = 8.820 J

Gravitational potential energy changes are;

ΔU(g) = U(gf) - U0

ΔU(g) = 8.820 J - 35.87 J

ΔU(g) = -27.05 J

Result, we can see that the work done through gravity is the opposite of how the gravitational potential energy has changed.

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