TS EAMCET 18-July-2022 Shift 2 Solved Paper

The given situation is shown below.

When the rocket will reach at maximum height, then the total gravitational potential energy at this height will be equal to the sum of kinetic energy at the surface of earth and the potential energy when the rocket was at the earth's surface.
∴‌‌−‌

GmMe

(R+h)

=‌

1

2

mv2+[‌

−GmMe

R

]
⇒‌‌‌

−GmMe

(R+h)

=‌

1

2

mv2−‌

GmMe

R

⇒‌‌‌

1

2

mv2=‌

GmMe

R

−‌

GmMe

R+h

‌⇒‌‌‌

1

2

v2=GMe[‌

1

R

−‌

1

R+h

]
‌⇒‌‌‌

1

2

×(4000)2=‌

GMc

R2

×R2[‌

R+h−R

R(R+h)

]‌. ‌
‌⇒‌‌8×106=gR2[‌

h

R(R+h)

]=‌

ghR

(R+h)

=‌

10hR

(R+h)

‌⇒‌‌.8×105=‌

hR

R+h

‌⇒8×105(R+h)=hR
‌⇒8×105[6.4×106+h]=h×6.4×106
‌⇒51.2×1011=(6.4×106−8×105)h
‌⇒51.2×1011=(64×105−8×105)h
⇒‌h=‌

51.2×1011

56×105

=0.914×106m
‌⇒‌‌h=‌

0.914×106

103

‌km=914‌km