Question

a freely falling body crosses the points a b c with velocities v,2v,3v.then AB:AC=​

Answer 1

Answer:

Let the acceleration be a.

for p to Q

intial speed u = v

final speed v = 2v

using v² - v³

2a Spq = ( 2v²) - v²

we get Spq = 3v²/2a

for Q to R

Intial speed v = 2v

final speed u = 3v

using v² - u² = 2aS

2aSqr = (3v)² -(2v)²

we get Sqr =Sv²/2a

so, Sqr : Spq = 3 :5

I HOPE THIS IS HELP FOR YOU