Question

A body of mass 5 kg is thrown vertically up with a kinetic energy of 1000 j. the height at which the kinetic energy of the body becomes half of the original value is ______.

Answer 1

Answer:

the answer is 5 metres.

))))) Step by step explination ((((((

Alright!

So, this is an issue of the conservation of energy, which states…

ΔEsys+ΔEelse=0ΔEsys+ΔEelse=0

Where ΔEΔE means change in energy, and sys and else refer to the system and the rest of the universe.

If we ignore air resistance, the ΔEelseΔEelse term becomes zero. So we get…

ΔEsys=0ΔEsys=0

We know that ΔE=ΔU+ΔKE+ΔPEΔE=ΔU+ΔKE+ΔPE

Where UU, KEKE, and PEPE are Internal Energy, Kinetic Energy, and Potential Energy respectively.

We will assume that the Internal Energy is not changing, which means…

ΔE=ΔKE+ΔPEΔE=ΔKE+ΔPE

Since ΔEΔE also equals 00

ΔKE+ΔPE=0→ΔKE=−ΔPEΔKE+ΔPE=0→ΔKE=−ΔPE

Now we’ve made all of our assumptions/simplifications, but there’s one last thing…

In a uniform gravitational field (which earth gravity mostly is for fairly low heights), you find that ΔPE=mgΔhΔPE=mgΔh where mm is mass, gg is gravitational field (acceleration due to gravity), and hh is change in height.

Now, we want to find ΔhΔh where ΔKE=−12KE0ΔKE=−12KE0 where KE0KE0 is some arbitrary initial value. So our final equation is this

−12KE0=−mgΔh→Δh=12mgKE0−12KE0=−mgΔh→Δh=12mgKE0

Now we just plug in all of the numbers for the specific situation the question asked for (which we know)

Δh=12mgKE0=12(5kg)(9.8ms2)(490J)=5mΔh=12mgKE0=12(5kg)(9.8ms2)(490J)=5m

HOPE IT HELPS...

PLEASE ,ARL AS BRAINLIEST..

PLEASE..