Question

Please help me with class 9 physics!!!

Answer 1

m = 2 kg
Let initial height be = h meters
potential Energy = m g h
velocity just before hitting the ground = V  m/s
Kinetic energy = 1/2 m V²
Energy is conserved. So   V² = 2 g h

Let velocity just after hitting the ground = v
KE = 1/2 m v²
Potential energy at the top position= m g (2 h / 5)
Then   v² = 4 gh / 5     ∵ conservation of energy

V² : v² = 2 g h / [ 4gh/5 ] = 5 / 2
V : v = √5 : √2
m V : m v  = √5 : √2           answer.

==

         KE = p² / (2 m) = PE = m g h
         p = √[2 m KE ] = m √[2 g h]

p1 : p2 = √h1 : √h2
            =  √h1 : √[ 2 h1/5 ]
            = √5 : √2

Answer 2

2 is ans...............