Question

A stone is thrown up from the surface of earth when it reaches at maximum height, its K.E is equal to :

Answer 1

Answer:

The KE of the stone is 0.

$kinetic energy = \frac{1}{2} mv {}^{2}$

Here v =0

Then

$\frac{1}{2?} \times m \times 0 \\ = 0$

Hence KE is 0

Answer 2

Given:

• Stone is thrown from surface of earth vertically up.

To find:

• Kinetic Energy at highest position ?

Calculation:

• Let's assume that the stone is thrown up with initial velocity of 'u'.

Now, this stone will reach a maximum height of u²/2g within a time of u/g.

• Once the stone reaches max height, it will temporarily be at rest (zero velocity).

• Then it will start to come downwards.

So, at max height, velocity is zero.

$KE_{max \: ht.} = \dfrac{1}{2} m {v}^{2}$

$\implies KE_{max \: ht.} = \dfrac{1}{2} m {(0)}^{2}$

$\implies KE_{max \: ht.} = 0 \: joule$

So, KE at max height is zero.