An airplane with a speed of 51.4 m/s is climbing upward at an angle of 48 An air

Question

An airplane with a speed of 51.4 m/s is climbing upward at an angle of 48

An airplane with a speed of 51.4 m/s is climbing upward at an angle of 48 degree with respect to the horizontal. When the plane's altitude is 700 m, the pilot releases a package. Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. Relative to the ground determine the angle of the velocity vector of the package just before impact. clockwise from the positive x axis

Explanation / Answer

a) y direction

y = y0 + v0y t + 1/2 at^2
0 = 700 + 51.4*sin(48)*t - 0.5*9.81*t^2

t=16.46 s
x dirextion
x = v0x t = 51.4*cos(48)*16.46=566.1 m

b) vx = 51.4*cos(48)
vy = 51.4*sin(48) - 9.81*16.46

angle = arctan vy/vx = arctan( (51.4*sin(48) - 9.81*16.46)/(51.4*cos(48)))=74.41 degrees

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