Posted by Corey Brown on .
A sniper fires a bullet at 120 m/s at 30° above the horizontal from the roof top of a 35 m high parking garage. If the bullet strikes the level ground beside the parking garage:
a) How long was the bullet in the air?
b) How far from the base of the parking garage did the bullet land?
c) At what angle did the bullet land?
Vo = 120m/s[30o]
Xo = 120*cos30 = 103.9 m/s.
Yo = 120*sin30 = 60 m/s.
a. Y = Yo * g*Tr = 0
Tr=-Yo/g = -60/-9.8 = 6.12 s. = Rise time.
h = ho + (Y^2-Yo^2)/2g
h = 35 + (0-(60^2))/-19.6 = 218.7 m.
h = 0.5g*t^2 = 218.7
4.9t^2 = 218.7
t^2 = 44.63
Tf = 6.68 s. = Fall time.
Tr+Tf = 6.12 + 6.68 = 12.8 s. = Time in
b. d = Xo*(Tr+Tf) = 103.9m/s * 12.8s =
c. Y^2 = Yo^2 + 2g*h
Y^2 = 0 + 19.6*218.7 = 4286.52
Y = 65.5 m/s.
Tan A = Y/Xo = 65.5/103.9 = 0.63041
A = 32.2o
TYPO: Change Y = Yo*g*Tr = 0 to
Y = Yo + g*Tr = 0
Thanks Henry -
You're a good guy!