An airplane with a speed of 89.3 m/s is climbing upward at an angle of 42.4 °?
with respect to the horizontal. When the plane's altitude is 899 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.
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Initial vertical velocity of the package is 89.3 sin 42.4 = - 80.26 m/s
Vertical displacement is 899 m
a =9.8 m/s^2
s = ut + ½ at^2
899 = - 80.26 t + 4.9t^2
t =24.02 s
Horizontal velocity is 89.3 cos 42.4 = 65.94 m/s
Horizontal distance traveled in this time is 24.02 *65.94 = 1583.98 m [Answer to (a) ]
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Final vertical velocity is v = - 80.26 + 9.8*24.02 Using v = u + at
v = 155.14 m/s
Resultant velocity is √ {155.14^2 +[ 89.3 cos 42.4]^2 } = 168.6 m/s
Tan θ = 155.14/ 65.94
θ = 66.97 ° below the horizontal .
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