a body of mass 5 kg is thrown vertically up with a kinetic energy of 690 Joel the height at which the kinetic energy of the body becomes half of the original value is
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Mass(m)=5 kg,acceleration due to gravity(g)=10 m/s²
Original energy=690 J
Required energy=1/2(690)=345 J
mgh=345 (E=mgh)
5×10×h=345
50 h=345⇒h=345/50=6.9 m(Ans.)
Original energy=690 J
Required energy=1/2(690)=345 J
mgh=345 (E=mgh)
5×10×h=345
50 h=345⇒h=345/50=6.9 m(Ans.)
Answered by
0
P E at ground = 0
and K= 690
At the top P.E=690 and K.E = 0 bcoz total energy remains constant
P.E=Mgh
690= 5*10*h
therefore: h= 13.8m so this is when P.E = maximum and K = 0
so at h\2 P.E is half = 13.8/2
= 6.9m
and K= 690
At the top P.E=690 and K.E = 0 bcoz total energy remains constant
P.E=Mgh
690= 5*10*h
therefore: h= 13.8m so this is when P.E = maximum and K = 0
so at h\2 P.E is half = 13.8/2
= 6.9m
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