Question

a ball is thrown with same speed 4m/s show that under free fall it with ball to the ground with same speed​

Answer 1

Answer:

let the speed with what the ball was thrown up be u

here ,

u= initial velocity

v= 0 (as ball will stop after a height )

let it's maximum height reached be s

so by the third equation of motion

v²-u²=2as(here a= -g as it is thrown up)

=>v²-u²=-2gs

=>-u²=-2gs(v=0)

=>u²=2gs

=>u=√2gs..........(1)

now the after reaching the maximum height the ball will start falling down

so,

it's initial speed will be 0 =u'

let it's final velocity be V'

therefore by the third equation of motion

v'²-u'²=2gs

=>v'²-u'²=2gs(u=0)

=>v=√2gs..........(2)

subtracting equation 1 from 2

we get ,

v'-u=√2gs-√2gs

=>v'-u=0

=>v'=u

therefore we can see that v'=u

so it will fall with the same speed as it was thrown up

.

hope it helps