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A Projectile IS FIRED FROM THE TOP OF A BUILDING WITH AN INITIAL VELOCITY STRAIGH UPWARD AT 40.0 M/SEC . THE TOP OF THE BUILDING IS 50.0 METERS ABOVE THE GROUND.
A) THE MAXIMUN HEIGHT OF THE PROJECTILE ABOVE THE BUILDING.
B) THE TIME FOR THE PROJECTILE TO REACH THE TOP.
C) THE VELOCITY OF THE PROJECTILE AS HIT THE GROUND
D) THE TOTAL TIME THE PROJECTILE IS IN THE AIR

PHYSICS 
Henry,
A. Vf^2 = Vo^2 + 2gh.
h = (Vf^2 _ Vo^2) / 2g,
h = (0  (40)^2) / 19.6 = 81.63m.
B. t(up) = (Vf  Vo) / g,
t(up) = (0  40) / 9.8 = 4.08s.
C. h = 50 + 81.63 = 131.63m. above ground.
Vf^2 = Vo^2 + 2g*h.
Vf^2 = 0 + 19.6*131.6 = 2580,
Vf = 50.8m/s.
D. t(dn) = (Vf  Vo) / g,
t(dn) = (50.8  0) / 9.8 = 5.18s.
T = t(up) + t(dn) = 4.05 + 5.18 = 9.23s
in air. 
PHYSICS 
Wisdom,
Maximum hieght is=46.9m
(B)answer=3.064m
time of flit is =6.128m