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Grade 9Mechanics

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

Prapul Muttineni

9 Years agoGrade 9

2 Answers

Vikas TU

9 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

7 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,

PE_{at point C}=mgh

KE_{at point C}=0

So, Total mechanical energy=PE_C+ KE_C=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,

TME_B=PE_B+KE_B

mgh =3mgh/4+KE_B

KE_B=mgh/4

Now,

KE_B:PE_B=mgh/4:3mgh/4

KE_B:PE_B=1:3

ANS:-1:3

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