Question

plss answer my question fast...........
with full solution ​

Answer 1

Answer :- option C ................

Answer 2

Answer:

    c. $s=(\frac{5}{6} )^2h$

Explanation:

Since, the ball is released from rest, its initial velocity is zero. When released from height h it reaches the ground in time T.

Applying equation of motion,we get,

$s=ut+\frac{1}{2}.gt^2\\\\ =>h=\frac{gt^2}{2} \\\\=>t^2=\frac{2h}{g}$

Now, we apply the same equation to calculate the height covered in time 5T/6

$=>s=ut+\frac{1}{2}at^2\\\\ =>s=\frac{g(\frac{5T}{6}^2 )}{2} \\\\=>s=\frac{25g}{72}*T^2\\\\ =>s=\frac{25g}{72}*\frac{2h}{g} \\\\=>s=\frac{25h}{36}\\\\=>s=(\frac{5}{6} )^2h$

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