A body is thrown vertically up with certain initial velocity. The potential and kinetic energies of the body are equal at a point P in its path. If the same body is thrown with double the velocity upwards, the ratio of potential and kinetic energies of the body when it crosses the same point, is

Question

A body is thrown vertically up with certain initial velocity. The potential and kinetic energies of the body are equal at a point P in its path. If the same body is thrown with double the velocity upwards, the ratio of potential and kinetic energies of the body when it crosses the same point, is (A) 1: 1 (B) 1: 4 (C) 1: 7 (D) 1: 8

Tardigrade Admin · Accepted Answer

Let a point where, (1/2) mv 2= mgh Total M.E =2 mgh Now, when v is twice Total M.E =8 mgh Now when the ball passes through that point K . E =8 mgh - mgh =7 mgh P ⋅ E=m g h P.E: K.E =1: 7

1 : 1

1 : 4

1 : 7

1 : 8

Let a point where, $\frac{1}{2} m v^{2} = m g h$
Total M.E $= 2 m g h$
Now, when v is twice
Total M.E $= 8 m g h$
Now when the ball passes through that point
$K . E = 8 m g h - m g h = 7 m g h$
$P \cdot E = m g h$
P.E : K.E $= 1 : 7$

Q. A body is thrown vertically up with certain initial velocity. The potential and kinetic energies of the body are equal at a point P in its path. If the same body is thrown with double the velocity upwards, the ratio of potential and kinetic energies of the body when it crosses the same point, is

1 : 1

1 : 4

1 : 7

1 : 8

Let a point where, 21​mv2=mgh Total M.E =2mgh Now, when v is twice Total M.E =8mgh Now when the ball passes through that point K.E=8mgh−mgh=7mgh P⋅E=mgh P.E : K.E =1:7

Let a point where, $\frac{1}{2} m v^{2} = m g h$
Total M.E $= 2 m g h$
Now, when v is twice
Total M.E $= 8 m g h$
Now when the ball passes through that point
$K . E = 8 m g h - m g h = 7 m g h$
$P \cdot E = m g h$
P.E : K.E $= 1 : 7$