Question

someone help me please​

Answer 1

Explanation:

it is of which classsssssssssssss

Answer 2

Explanation:

Consider an object of mass m is thrown upwards to a height h above the ground level.

Taking different point one by one.

At point A,

The body is at rest and v = 0

So, PE=mgh and KE=0

So, total energy,

TE=PE+KE

TE=mgh..............(a)

At point B,

PE=mg(h−x)=mgh−mgx...........(1)

KE=

2

1

mv

2

Using equation of motion as

v

2

−u

2

=2as

∵u=0

v

2

=2gx

so,KE=

2

1

m(2gx)=mgx

Total energy,

TE=mgh−mgx+mgx

TE=mgh.........(b)

At point C:

The body hits the ground so, u = 0 and h = 0. So, it possess only kinetic energy

KE=

2

1

mv

2

v

2

=u

2

+2as

v

2

=2gh

so,

KE=

2

1

m(2gh)

KE=mgh

so,

TE=mgh.........(c)

From equation (a), (b) and (c) the total energy at point A, B and C is same i.e. mgh.

TE∝h

So, the above is the correct graph for TE vs h.