Posted by HELPPMEEEEEEEEEE on Sunday, October 16, 2011 at 5:11pm.
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, Monday, October 17, 2011 at 4:12pm
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, Friday, March 4, 2016 at 3:31pm
Maximum hieght is=46.9m
(B)answer=3.064m
time of flit is =6.128m
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