Monday

October 20, 2014

October 20, 2014

Posted by **HELPPMEEEEEEEEEE** on Sunday, October 16, 2011 at 5:11pm.

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 4:12pmA. 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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