A stone is projected vertically up with a velocity U, reaches upto a maximum height h. When it is at a height of (3 h/4) from the ground, the ratio of KE and PE at that point is: (.PE=0 at the point of projection)

Question

A stone is projected vertically up with a velocity U, reaches upto a maximum height h. When it is at a height of (3 h/4) from the ground, the ratio of KE and PE at that point is: (.PE=0 at the point of projection) (A) 1:1 (B) 1:2 (C) 1:3 (D) 3:1

Tardigrade Admin · Accepted Answer

(1/2)mu2=mgh at a height (3 h/4)P.E. =mg((3 h/4)) K.E=mg((h/4)) KE: pot. energy 1:3

Q. A stone is projected vertically up with a velocity $U,$ reaches upto a maximum height h. When it is at a height of $\frac{3 h}{4}$ from the ground, the ratio of $K E$ and $PE$ at that point is : $(PE = 0$ at the point of projection)

$1 : 1$

$1 : 2$

$1 : 3$

$3 : 1$

$\frac{1}{2} m u^{2} = m g h$
at a height $\frac{3 h}{4} P . E .$ $= m g (\frac{3 h}{4})$
$K . E = m g (\frac{h}{4})$
$K E :$ pot. energy
$1 : 3$

Q. A stone is projected vertically up with a velocity U, reaches upto a maximum height h. When it is at a height of 43h​ from the ground, the ratio of KE and PE at that point is : (PE=0 at the point of projection)

1:1

1:2

1:3

3:1