50 points....
class 11
....
physics...

Answers
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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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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