$(a)$ How much momentum will an object of mass $10 \ kg$ transfer to the floor if it falls from a height of $0.8 \ m$? $(g = 10 \ m s^{-2})$
$(b)$ Explain why it is difficult for a fireman to hold a hose,which ejects a large amount of water at a high velocity.

(N/A) Given: Mass $m = 10 \ kg$,initial velocity $u = 0 \ m s^{-1}$,acceleration $a = g = 10 \ m s^{-2}$,and displacement $S = 0.8 \ m$.
Using the equation of motion $v^2 - u^2 = 2aS$:
$v^2 - 0^2 = 2 \times 10 \times 0.8$
$v^2 = 16$
$v = 4 \ m s^{-1}$
Now,momentum $p = m \times v = 10 \ kg \times 4 \ m s^{-1} = 40 \ kg \ m s^{-1}$.
$(b)$ According to Newton's third law of motion,for every action,there is an equal and opposite reaction. When water is ejected from the hose at a high velocity,it exerts a forward force (action). The hose,in turn,exerts an equal and opposite backward force (reaction) on the fireman. To counteract this reaction force,the fireman must hold the hose with a significant amount of force,making it difficult to control.