Question

2.2JJ
A ball of mass 50 g is thrown upwards. It rises to a maximum height of 100 m. At what height its
K.E. will be reduced to 70%.
(B) 40 m
(D) 70 m
(A) 30 m
(C) 60 m
by 10cm​

Answer 1

Answer:

ANSWER

v^2=u^2+2as

0=u^2+2(−10)(100)

u= 2000=20^5

KE initially= 0.5mu^2

=0.5(0.05)(2000)=50J

0.7KE. = 0.7(50)= 35J= 0.5mv^2

v=2KE/m

=2(35)/0.05

= 1400

v^2=u^2+2as

1400=2000+2(-10)s

s= 30m

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