# Mass of a body is 500g . It is thrown vertically upward with kinetic energy 58.8j . So its kinetic energy reduce to half at which height?

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Gaya3 Share
Oct 21, 2017

$6 \text{m}$

#### Explanation:

Initial Kinetic Energy $= {\text{KE}}_{i} = 58.8 J$

Final Kinetic Energy $= {\text{KE}}_{f} = \frac{58.8}{2} = 29.4 J$

Let's assume that the rest of Kinetic Energy has been converted into Gravitational potential energy.

Initial Gravitational potential Energy $= {\text{PE}}_{i} = 0 J$

Final Gravitational potential Energy $= {\text{PE}}_{f} = 58.8 - 29.4 = 29.4 J$

We need to know the height $\textcolor{b l u e}{h}$

So, ${\text{PE}}_{f} = m g \textcolor{b l u e}{h}$

$\implies 29.4 J = 0.5 \text{kg} \times 9.8 \frac{m}{s} ^ 2 \times \textcolor{b l u e}{h}$

$\implies \textcolor{b l u e}{h} = \frac{29.4 J}{0.5 \text{kg} \times 9.8 \frac{m}{s} ^ 2}$

$\implies \textcolor{b l u e}{h} = \frac{29.4 \cancel{N} m}{0.5 \cancel{\text{kg}} \times 9.8 \cancel{\frac{m}{s} ^ 2}}$

$\textcolor{b l u e}{h} = 6 m$

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Oct 21, 2017

6m

#### Explanation:

When the Kinetic energy has reduced to half, the other part has become gravitational potential energy ${E}_{w} = m g h = \frac{58 , 8 J}{2} = 29 , 4 J$
hence $h = {E}_{w} / \left(m g\right) = \frac{29 , 4 J}{0 , 5 k g \times 9 , 8 \frac{m}{s} ^ 2} = 6 m$

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