A man throws the brick to a height of 12 m with a speed of 12 m/s. If he throws the brick in such a way that they just reach that height , what percentage of energy will be saved

$(a)\;29\;\% \quad (b)\;38\;\% \quad (c)\;46\;\% \quad (d)\;50\;\%$

When the brick reaches 12 m height when thrown  with velocity 12 m/s it has both potential energy and kinetic energy
When it just reaches the same height, it has only Potential energy, as its velocity is zero at its maximum height.
Therefore $\%$ of energy saved $=\large\frac{KE}{KE+PE}$$\times 100 \qquad= \large\frac{\Large\frac{1}{2}mv^2}{\Large\frac{1}{2}mv^2+mgh}$$\times 100$
$\qquad=\large\frac{12^2}{12^2+2 \times 9.8 \times 12}$$\times 100$
$\qquad=38 \%$
Hence  b is the correct answer.

edited Feb 10, 2014 by meena.p