Question

if we throw a body vertically upward with velocity of 4 metre per second then at what height does its kinetic energy reduced to half of the initial value?

Answer 1

this solution should do. only equations of k.e is needed

Answer 2

At 0.408 meters kinetic energy is reduced to half of the initial value.

Explanation:

Let $K_i\ and\ K_f$ are the initial and the final kinetic energy of a body. The initial kinetic energy is given by :

$K_i=\dfrac{1}{2}mv^2$

Speed with which the ball is thrown upward, v = 4 m/s

Let at height h the kinetic energy reduced to half of the initial value. So,

$K_f=\dfrac{1}{2}K_i$

$\dfrac{1}{2}mv'^2=\dfrac{1}{2}\times \dfrac{1}{2}mv^2$.............(1)

When it reaches to a height h, using the conservation of energy as :

$\dfrac{1}{2}mv'^2=mgh$

$v'^2=2gh$

So, equation (1) becomes :

$2gh=\dfrac{1}{2}(4)^2$

h = 0.408 meters

So, at 0.408 meters kinetic energy is reduced to half of the initial value. Hence, this is the required solution.

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Conservation of energy

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