$A$ piece of stone is thrown vertically upwards. It reaches its maximum height in $3 \ s$. If the acceleration of the stone is $9.8 \ m s^{-2}$ directed towards the ground,calculate the initial velocity of the stone with which it is thrown upwards. Find the maximum height attained by it.

(A) Given: Time $t = 3 \ s$,Final velocity $v = 0 \ m s^{-1}$ (at maximum height),Acceleration $a = g = -9.8 \ m s^{-2}$ (acting downwards).
To find initial velocity $u$:
Using the first equation of motion: $v = u + at$
$0 = u + (-9.8) \times 3$
$u = 29.4 \ m s^{-1}$.
To find maximum height $h$:
Using the second equation of motion: $h = ut + \frac{1}{2}at^2$
$h = (29.4 \times 3) + \frac{1}{2} \times (-9.8) \times (3)^2$
$h = 88.2 - 44.1$
$h = 44.1 \ m$.