$A$ ball thrown up vertically returns to the thrower after $6 \, s$. Find:
$(a)$ The velocity with which it was thrown up.
$(b)$ The maximum height attained. (Take $g = 9.8 \, m s^{-2}$)

(N/A) Given: Total time $T = 6 \, s$,final velocity at maximum height $v = 0 \, m s^{-1}$,acceleration due to gravity $g = 9.8 \, m s^{-2}$.
Since the time of ascent equals the time of descent,the time taken to reach the maximum height is $t = T / 2 = 6 / 2 = 3 \, s$.
$(a)$ Using the first equation of motion,$v = u + at$,where $a = -g$:
$0 = u + (-9.8 \times 3)$
$u = 29.4 \, m s^{-1}$.
Thus,the initial velocity is $29.4 \, m s^{-1}$.
$(b)$ Using the third equation of motion,$v^2 - u^2 = 2as$:
$0^2 - (29.4)^2 = 2 \times (-9.8) \times h$
$-864.36 = -19.6 \times h$
$h = 864.36 / 19.6 = 44.1 \, m$.
Thus,the maximum height attained is $44.1 \, m$.