Question

a boy throws a ball of mass 20g vertically upwards .It rises to max. height of 100m. At what height will K.E. be reduced to 70%

Answer 1

let ball throws with u m/s .
at maximum height velocity of ball =0

use formula,
v^2 = u^2 + 2aS
0 = u^2 -2(g)(100)
where g is acceleration due to gravity.

u = 10√(2g) m/s

now,
initail K.E = 1/2mu^2 =1/2(20/1000) (200g) =2g Joule

if K.E reduced to 70%
then,
new K.E =2g x 70/100 = (1.4)g

1/2mv^2 = (1.4)g
1/2(0.02) × v^2 = (1.4)g
(0.01) × v^2 = (1.4)g
v^2 = 140g

now use ,
v^2 = u^2 +2aS
140g =200g -2gH

H = 30m