1. An airplane with a speed of 92.9 m/s is climbing upward at an angle of 60.4 °
ID: 1416248 • Letter: 1
Question
1. An airplane with a speed of 92.9 m/s is climbing upward at an angle of 60.4 ° with respect to the horizontal. When the plane's altitude is 986 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.
2.An eagle is flying horizontally at 8.2 m/s with a fish in its claws. It accidentally drops the fish. (a) How much time passes before the fish's speed doubles? (b) How much additional time would be required for the speed to double again?
Explanation / Answer
vox = vo cos 60.4 = 92.9 cos60.4 = 45.88 m/s
voy = 92.9 sin60.4 = 80.77 m/s
from the kinematic equation
s= voy t+ 1/2 at^2
-986 m = 80.77 m/s * t - 1/2 ( 9.8) t^2
4.9 t^2 - 80.77 t -986 m = 0
solving quadratic equation
t = 24.64 s
Range ( horizontal distance ) = x= vox t = 45.88 ( 24.64 s) = 1130.48 m
(b)
vy = voy-gt = 80.77- 9.8(24.64) = -160.7 m/s
The final velocity
v = sqrt vox^2 + vy^2 = sqrt ( 45.88)^2 + ( -160.7)^2 = 167.12 m/s
The angle made by the package relative to the ground is
= cos^-1(45.88m/s /167.12 m/s)
= 74 degree
As per guide liens I worked first problem, please post reamaining question next time
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