Question

50 points....


class 11
....

physics...​

Answer 1

Answer:

maximum height attained be h

and total energy at height h = mgh

Let at height H, the kinectic energy will be 75%

Totao mechanical energy at height H be

0.75 KE + mgH

as KE at maximum height = total energy = mgh

put the value of KE in above eqn

0.75 mgh + mgH

now equating the total energy we get

0.75 mgh + mgH = mgh

or mgH = 0.25 mgh

or H = 0.25 h

therefore at this height the KE becomes 75%.

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Answer 2

Hello friends.

Let vf = final velocity

Vi = initial velocity.

Acc to ques

K.Efinal=3/4 K.Einitial (75% of initial)

1/2mvf^2 = 3/4(1/2)mvi^2

vf^2 = (3/4)vi^2…….. eqn 1

Now if H is the maximum height achieved by the ball when vf=0 therefore,

vf^2 = vi^2 −2gs

0=vi^2 − 2gH

vi^2 = 2gH ………… eqn 2

Also for the height h when the K.E. is 75% it's maximum, we use the kinematic equation using vf obtained earlier in eqn 1

(3/4)vi^2 = vi^2 −2gh

2gh=(1/4)vi^2

Putting equation 2 here -

2gh = (1/4)2gH

h=H/4.

hope it's helpful for you..

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