Question

A labour has to throw bricks near mistry 16 feet vertically above. He throws the bricks in such a manner that the brick reach the mistry with the velocity of 16ft/sec. If he throws bricks such that bricks just reach the mistry, then the portion of the energy saved is ?

Answer 1

portion of the energy saved is  1/5

Explanation:

a = -g

using V² - U²  = 2aS

V = 16 ft/s

=> 16² - U² = 2(-g)(16)

=> U²  = 16²  + 2(g)(16)

Energy required = (1/2)mU²   = (1/2)m(16²  + 2(g)(16))

using g = 32 ft/s

= (1/2)m(16²  + 2(32)(16))

= (1/2)m(16)²(1 + 4)

= (1/2)m(16)²(5)

throws bricks such that bricks just reach the mistry,

=> Final Velocity is Zero

using V² - U²  = 2aS

=> 0² - U² = 2(-g)(16)

=> U²  = 2(g)(16)

Energy required = (1/2)mU²   = (1/2)m(2(32)(16))

= (1/2)m(16)²(4)

Energy Saved = ((1/2)m(16)²(5)  -  (1/2)m(16)²(4))/  (1/2)m(16)²(5)

= (1/2)m(16)²(1)/  (1/2)m(16)²(5)

= 1/5

1/5th portion of the energy is saved

Another way

Energy at top in 1st case  =  mgh + (1/2)mv²  = m(32)(16) + (1/2)m16² = (1/2)m(16)²(5)

& Energy at top  in  2nd case =  mgh + 0  = m(32)(16)  = (1/2)m(16)²(4)

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Answer 2

$1/5$

Portion used is 1/5