Question

find the height at which K.E. is 1/4 of the P.E. , when a mass is dropped from a heght of 20m

Answer 1

K.E=P.E/4

at maximam height P.E is max
K.E=0

let x be the height
then
V²=2gx
K.E=1/2 mv²
=1/2 *m*2gx=mgx
P.E=mgh
=mg*(20-x)
=20mg-mgx

given mg*(20-x)/4=mgx
mg(20-x)= 4mgx
g(20-x)=4gx
9.8*20-9.8x=39.2x
196=39.2x+9.8x=49x
x=196/49=4 m

hope it helps
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Answer 2

K.E=P.E/4

at maximam height P.E is max & K.E=0

let x be the height

then

V²=2gx

K.E=1/2 mv²

=1/2 ×m×2gx=mgx

P.E=mgh

=mg×(20-x)

=20mg-mgx

GIVEN

mg×(20-x)/4=mgx

⇒mg(20-x)= 4mgx

⇒g(20-x)=4gx

⇒9.8×20-9.8x=39.2x

⇒196=39.2x+9.8x=49x

⇒x=196/49=4 m