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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

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

Grade:9

2 Answers

Vikas TU
14149 Points
7 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/4
Ratio => K.E./P.E. = (0.5u^2 + 3gh/4 )/(3gh/4)       
Raghav
13 Points
5 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 =mgh
KEat point C =0
So, Total mechanical energy=PEC + KEC=mgh
Since 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+KEB
mgh  =3mgh/4+KEB
KEB =mgh/4
Now,
KEB:PEB=mgh/4:3mgh/4
KEB:PEB=1:3
ANS:-1:3
             
           
 

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