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`        A stone is’projected up with a velocity u reaches a max height h.When it is at a height of 3h/4 from the ground,the ratio of K.E and P.E is?(consider P.E =0 at the point of projectory`
2 years ago

```							P.E = mg(3h/4)And K.E. = 0.5mv^2 Now, v at 3h/4,v = root(u^2 + 3gh/2)K.E. = 0.5mu^2 + 3mgh/4Ratio => K.E./P.E. = (0.5u^2 + 3gh/4 )/(3gh/4)
```
2 years ago
```							Let the point from where the stone is projected be A. Let the point where the stone attains the maximum height 'h' be C. Let the point which is at a height of 3h/4 from the feline be B. Let the mass of stone be 'm'Now,PEat point C =mghKEat point C =0So, Total mechanical energy=PEC + KEC=mghSince the only force acting on the stone is gravitational force which is a conserved force so the total mechanical energy at every point in the system will always be constant.So,TMEB=PEB+KEBmgh  =3mgh/4+KEBKEB =mgh/4Now,KEB:PEB=mgh/4:3mgh/4KEB:PEB=1:3ANS:-1:3
```
one year ago
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