Question

Tu
79.
Maximum height reached by an object projected
perpendicular to the surface of the earth with
speed equal to 25% of the escape velocity from
earth surface is - (R = Radius of Earth) :-
R.
16R
R
8R
(2)
(3)
(4)
9
15
7​

Answer 1

Explanation:

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Answer 2

The maximum height reached by an object is R/15.

Given:

An object is projected perpendicular to the surface of the earth with a speed equal to 25% of the escape velocity from the earth's surface.

To find:

The maximum height reached by an object

Solution:

By using the conservation of energy principle.

The escape velocity from the surface of the earth

                         V$_{e}$ = √(2GM/R)  

Here G = the gravitational constant,

M = Mass of the earth,  

R = Radius of the earth.

If the initial speed of the object is 25% of the escape velocity,

then its initial kinetic energy will be:

K = (1/2)mv² = (1/2)m(0.25V$_{e}$)² = 0.125m(V$_{e}$)²

At the maximum height, the object's final velocity will be zero, and its initial kinetic energy will be converted to potential energy.

The potential energy at the maximum height is given by:

                            U = mgh

where h is the maximum height.

Using the conservation of energy principle, we can equate the initial kinetic energy to the potential energy at the maximum height:

=> K = U

=> 0.03125m(V$_{e}$)² = mgh

Now solve it for h,  

h = (0.03125(V$_{e}$)²)/(g)

Substituting  V$_{e}$ = √(2GM/R) and g = GM/R²,

h = (0.03125(√(2GM/R))²/ GM/R²

h = (0.03125(2GM/R)/ GM/R²  

h = 0.0625R ≅ R/15

Therefore,

The maximum height reached by an object is R/15.

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