Question

a body of mass 1kg drops from the top of tower of height 50m what will be its K.E 10m below the top​

Answer 1

Answer:

Assuming that there is no air resistance (no heat energy produced) then the loss in gravitational potential energy (mgh) equals the gain in kinetic energy.

The kinetic energy at the top is zero (as there is no motion).

So, kinetic energy after falling 10 metres = m g h

= 1 x 9.81 x 10

= 98.1 J

In practice, there will be air resistance so the kinetic energy will be less than this answer.

Answer 2

GiveN :-

• A body of mass 1kg drops from the top of tower of height 50m .

To FinD :-

• The K.E 10m below the top .

SolutioN :-

Given that a body of mass 1kg is dropped from a tower of height 50m . We need to find the the kinetic energy 10m below the top. So ,

$\red{\frak{ Given }}\begin{cases} \textsf{ Mass of the body =\textbf{ 1 kg } . } \\\textsf{ Height of the tower =\textbf{ 50 m } . }\end{cases}$

We will have to find its velocity at 10m below from the top . Which can be calculated from the formula $\bf v_{( At\ h )}=\sqrt{ 2gh }$ which is applicable under free fall .

★ Using the formula of Kinetic Energy:-

$\sf:\implies\pink{ K.E . =\dfrac{1}{2} mv^2}\\\\\sf:\implies K.E. =\dfrac{1}{2}\times (1kg)^2 \times (\sqrt{2gh })^2 \\\\\sf:\implies K.E. =\dfrac{1}{2} \times 1 \times 2\times 10 \times 10 \\\\\sf:\implies \underset{\blue{\bf Required\ Energy}}{\underbrace{\boxed{\pink{\frak{ K.E. = 100 \ Joules }}}}}$