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An airplane with a speed of 13 m/s is climbing upward at an angle of 48° counterclockwise from the positive x axis. When the pl...Asked by Prachi
An airplane with a speed of 27.4 m/s is climbing upward at an angle of 48° counterclockwise from the positive x axis. When the plane's altitude is 860 m the pilot releases a package.
(a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.
Determine the angle of the velocity vector of the package just before impact.
(a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.
Determine the angle of the velocity vector of the package just before impact.
Answers
Answered by
Henry
Vo = 27.4m/s[48o]
Xo = 27.4*Cos48 = 18.3 m/s.
Yo = 27.4*sin48 = 20.4 m/s.
h = 0.5g*t^2 = 860 m.
4.9*t^2 = 860
t^2 = 175.5
Tf = 13.2 s. = Fall time.
Dx = Xo*Tf = 18.3m/s * 13.2s = 242.4 m.
b. Y = Yo + g*t = 0 + 9.8*13.2 = 129.4
m/s.
Tan A = Y/Xo = 129.4/18.3 = 7.07104
A = 82o.
Xo = 27.4*Cos48 = 18.3 m/s.
Yo = 27.4*sin48 = 20.4 m/s.
h = 0.5g*t^2 = 860 m.
4.9*t^2 = 860
t^2 = 175.5
Tf = 13.2 s. = Fall time.
Dx = Xo*Tf = 18.3m/s * 13.2s = 242.4 m.
b. Y = Yo + g*t = 0 + 9.8*13.2 = 129.4
m/s.
Tan A = Y/Xo = 129.4/18.3 = 7.07104
A = 82o.