$A$ 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) Given: Mass $m = 0.5 \, kg$,initial velocity $u = 25 \, m s^{-1}$,acceleration due to gravity $g = -10 \, m s^{-2}$.
$(a)$ Initial momentum $p = m \times u = 0.5 \times 25 = 12.5 \, kg \, m s^{-1}$.
$(b)$ To find the momentum at half the maximum height,first calculate the maximum height $H$ using $v^2 - u^2 = 2gH$. At maximum height,final velocity $v = 0$.
$0^2 - (25)^2 = 2 \times (-10) \times H$
$-625 = -20 \times H$
$H = 625 / 20 = 31.25 \, m$.
Halfway height $h = H / 2 = 31.25 / 2 = 15.625 \, m$.
Now,find the velocity $v_h$ at height $h$ using $v_h^2 - u^2 = 2gh$:
$v_h^2 - (25)^2 = 2 \times (-10) \times 15.625$
$v_h^2 - 625 = -312.5$
$v_h^2 = 312.5$
$v_h = \sqrt{312.5} \approx 17.68 \, m s^{-1}$.
Momentum at halfway mark $p_h = m \times v_h = 0.5 \times 17.68 = 8.84 \, kg \, m s^{-1}$.