Question

A ball is falling freely from a certain height of h when it reaches 10m height from the ground its velocity v is v'.It collides with horizontal ground and loses 50% of energy and rises back to a height of 10m.The value of v' is

Answer 1

HEY BUDDY HERE IS UR ANSWER
Height h = 10m
Initial velocity = Vo
Let the mass of ball be m
The ball is projected downwards the with Vo from height h
Total energy= mgh+1/2 mvo^2
total energy after the collision= 50/100 × (mgh)+1/2mvo^2
the ball rises back to the same height h after collision hence,
vo= √2gh=√2*9.8*10=√20*9.8=14m/s

Answer 2

Answer:

The velocity of ball when it reaches height 10m from the ground v' is equal to $10\sqrt{2}ms^{-1}$

Explanation:

Consider that the velocity v₀ to rise the ball up to height 10m.

Use the law of conservation of energy

$K.E.=P.E.$

$\frac{1}{2}mv^{2}_{o}=mgh$

$v_{o}=\sqrt{2gh} = \sqrt{(2)(10)(10)}$                      [g≈ 10ms⁻²]

$v_{o}=10\sqrt{2} ms^{-1}$

The kinetic energy will be doubled before the collision than that of after the collision.

$K.E_{Before} =2*K.E_{after}$

$\frac{1}{2}mv^{2}=2(\frac{1}{2}mv^{2}_{o} )$

$v=v_{o}\sqrt{2} \\ v=(10\sqrt{2})\sqrt{2}=20ms^{-1}$

Velocity of collision is $v=20ms^{-1}$

By using 3rd equation of motion:

$v^{2}-v'^{2}=2aS$

$(20)^{2}-v'^{2}=2(10)(10)\\ v'^{2}=400-200\\ v'=\sqrt{200}\\ v'=10\sqrt{2} ms^{-1}$

Therefore the velocity of the given ball is 10√2ms⁻¹ at height 10m from the ground.