Question

stone thrown upward X3 second to attain maximum height calculate first initial velocity of stone second maximum height attained by stone third net displacement when it come back

Answer 1

$\Longrightarrow{\boxed{\bold{Answer: 30\: ms^{-1}\:and\:0}}}$

$\textbf{\underline{Step-by-step explanation :- }}$

➣ When stone is thrown from down to up then at some time after reaching its maximum height, its velocity will come back to 0.

According to 1s equation of motion,

$\textbf{\huge{v = u + gt }}$

➣ Since ball is thrown against gravity that's why g = - 10 m/s²

0 = u + ( - 10)(3)

0 = u - 30 m/s

$\bold{\huge{30\:ms^{-1}\:=\:u}}$

➣ When ball comes back to us then its initial position is same as before.

➣ Since displacement is all about final position - initial position so answer will be 0 because here final and initial position are same.