time in air:
hf=h+355*sin68.7*t - 4.9t^2
t= 355sin68.7/4.9=67.5sec
dhoriztal=355*67.5*(1-.321)
A projectile is fired at v0= 355.0 m/s at an angle of 68.7% with respect to the horizontal. Assume that air friction will shorten the range by 32.1%. How far will the projectile travel in the horizontal direction, R?
2 answers
Vo = 355m/s[68.7o].
Xo = 355*Cos68.7 = 129.0 m/s.
Yo = 355*sin68.7 = 330.8 m/s.
Y = Yo + g*Tr = 0,
330.8 + (-9.8)Tr = 0,
Tr = 33.80 s. = Rise time.
Tf = Tr = 33..80 s. = Fall time.
R = Xo*(Tr+Tf) = 129*67.6 = 8720.4 m. with no air friction.
R = 8720.4 - 0.321*8720.4 = 5921 m. with air friction.
Xo = 355*Cos68.7 = 129.0 m/s.
Yo = 355*sin68.7 = 330.8 m/s.
Y = Yo + g*Tr = 0,
330.8 + (-9.8)Tr = 0,
Tr = 33.80 s. = Rise time.
Tf = Tr = 33..80 s. = Fall time.
R = Xo*(Tr+Tf) = 129*67.6 = 8720.4 m. with no air friction.
R = 8720.4 - 0.321*8720.4 = 5921 m. with air friction.
2 answers