$A$ man throws a ball of mass $0.5\, kg$ vertically upward with a velocity of $25\, m s^{-1}$. Find:
$(a)$ The initial momentum of the ball.
$(b)$ The momentum of the ball at the halfway mark of the maximum height (given $g = 10\, m s^{-2}$).

(N/A) Initial momentum is calculated as $p = m \times u = 0.5 \times 25 = 12.5\, kg\, m s^{-1}$.
$(b)$ First,find the maximum height $h$ using the equation $v^2 = u^2 - 2gh$. At maximum height,$v = 0$,so $0 = (25)^2 - 2 \times 10 \times h$,which gives $h = 625 / 20 = 31.25\, m$.
The halfway mark is $H = h / 2 = 31.25 / 2 = 15.625\, m$.
Now,find the velocity $v$ at this height using $v^2 = u^2 - 2gH$:
$v^2 = (25)^2 - 2 \times 10 \times 15.625 = 625 - 312.5 = 312.5$.
$v = \sqrt{312.5} \approx 17.68\, m s^{-1}$.
The momentum at the halfway mark is $p = m \times v = 0.5 \times 17.68 = 8.84\, kg\, m s^{-1}$.