An airplane with a speed of 99.8 m/s is climbing upward at an angle of 43.2 An a

Question

An airplane with a speed of 99.8 m/s is climbing upward at an angle of 43.2

An airplane with a speed of 99.8 m/s is climbing upward at an angle of 43.2 degree with respect to the horizontal. When the plane's altitude is 717 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. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

Explanation / Answer

Vx = 99.8*cos43.2 = 72.751 m/s

a) -h = Xtan43.2 - 0.5*g*X^2/Vx^2

-717 = X*0.939 - 4.9*X^2/72.751^2

X = 1522.826 m

b) Vy just befor impact
Vyf^2 - Vyi^2 = 2*-g*-h

Vyf = 136.823 m/s

angle = tan^-1(Vyf/Vx) = 62.03

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