Question

The PE and KE of a helicopter flying horizontally at a height of 400 m are in the ratio of \[5:2\]. The velocity of the helicopter is:
(A) \[28{\text{ }}m/s\]
(B) \[14{\text{ }}m/s\]
(C) \[56{\text{ }}m/s\]
(D) \[30{\text{ }}m/s\]

Answer 1

Hint We need to calculate the potential energy and kinetic energy of the helicopter and then divide them. That ratio will be equal to \[5:2\], except for the velocity all other quantities are known.

Complete step by step answer
Potential energy of a body at a height h is given as:
\[PE\, = \,mgh\]
Where m is the mass of the body
g is the acceleration due to gravity
h is the height of the body above the surface of the earth
Also, kinetic energy of the body is given as
\[KE\, = \,\dfrac{{m{v^2}}}{2}\]
Where m is the mass of the body
v is the velocity of the body
The ratio of these 2 quantities is given as \[5:2\], equating
\[
\dfrac{{mgh}}{{\dfrac{1}{2}m{v^2}}}\, = \,\dfrac{5}{2} \\
\Rightarrow \dfrac{{2gh}}{{{v^2}}}\, = \,\dfrac{5}{2} \\
\Rightarrow \dfrac{{4\,\times\,9.81\,\times\,400}}{5}\, = \,{v^2} \\
\Rightarrow v = \sqrt {\dfrac{{4\,\times\,9.81\,\times\,400}}{5}} \\
\Rightarrow v\, = \,56.7 \\
\]

Therefore the option with the correct answer is option C.

Note Make sure of the ratios. Usually, options also mention the reciprocals in the answers.
Make sure that you only use mgh when the body is nearly on the surface of the earth else one can go wrong.
Potential energy between 2 bodies is also written as \[PE{\text{ }} = {\text{ }} - \dfrac{{GMm}}{r}\], where M and m are the masses of 2 bodies, this is valid everywhere, however the equation mgh is valid only for a body resting on earth or very close to the surface of Earth.