Question

A ball,initially at rest is released from the top of a tower of height of 27 m .It takes 'T'seconds to reach the ground .Find the height of the ball ,above the ground at t =T/3

Answer 1

the question was very tricky

Answer 2

Hey

$Distance \ from \ Ground = S_G \\ Distance \ from \ top \ to \ height \ at \ 't= ( T / 3 )' = S_t \\ \\ S_t = 0.5 g \ (T/3)^2 \\ S_G = 0.5g \ T^2 \\ \\ Hence \ height \ of \ ball \ from \ Ground \ after \ ( T/3 )s = [ S_G - S_t ] \\ = [ \frac{4}{9} g \ T^2 ]$

And that's your answer ^_^  [ 4/9 g T² ] --> ( i )

See, since we already knew about time 'T' and 'T/3' , we are only applying the formula --> s = ut + 0.5 at² √√

$Since, S_g = ( 1/2 )g \ T^2 = 27m \\ \\ We \ put \ this \ in \ Eqn. ( i ) \ to \ get, \\ \\ --\ \textgreater \ Answer = \frac{4}{9} g \ T^2 = \frac{8}{9} (1/2)g \ T^2 = \frac{8}{9} * 27 = 24m$

Hence, your final answer = 24m

Hope this helped ^_^