Question

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

Answer 1

at 5.09 m,/sec*sec hope the answer help you

Answer 2

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