a body of mass 10 kg is thrown vertically with a KE 500 joules if the acceleration is 10 m/sec×sec the height at which the KE of the body becomes half of original value
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at 5.09 m,/sec*sec hope the answer help you
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TE=KE+PE
so, initially
PE=0 (h=0)
so 1/2mv2=500
on solving we get v=10 m/s
now, when the KE is becoming half the original value, the PE is equal to KE,
TE is conserved (using conservation of energy)
so PE=1/2KE
mgh=1/2(1/2mv2)
therefore, h=2.5 m
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