The projectile (ball) reaches an instantaneous vertical speed (Vy) of zero at maximum height.
so, V(max height) = ¬Г(Vx)^2+(Vy)^2
in this case V(max height) = Vx, where Vy=0
The maximum height, Yf, can be solved using Vfy^2=Viy^2 + 2gy. At maximum height Vfy=0.
Answer:

Explanation:
Pressure at the bottom of the building and at the top of the building must be related as



now we will have




now we have


A force is a push or pull to an object
Answer:
if im not mistaken i think its d let me know if correct plz