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.

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