Asked by Corey Brown

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?

Answers

Answered by Henry
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.
Above gnd.

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
air.

b. d = Xo*(Tr+Tf) = 103.9m/s * 12.8s =
1330 m.

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
Answered by Henry
TYPO: Change Y = Yo*g*Tr = 0 to
Y = Yo + g*Tr = 0
Answered by Corey Brown
Thanks Henry,

You're a good guy!
Answered by henry the best
honestly henry u is smart and loyal love ya
Answered by Aman Dixit
What is Y supposed to be?
Answered by Aman Dixit
Y^2 = Yo^2 + 2g*h

In that formula, what is Y??

I know Yo is vertical velocity, g is gravity and h is height but what is Y?
Answer
DAMN! they teaching kids about rifle basics in school?!?!?! WOW, just wow.
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