Calculating Distance and Velocity in Two-Dimensional Kinematics

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SUMMARY

The discussion focuses on calculating the distance and velocity of a package released from an airplane climbing at an angle of 34.1° with a speed of 77.5 m/s. The altitude at the time of release is 585 m. The calculated horizontal distance to the impact point is 1040 m, and the angle of the velocity vector just before impact is -61.0°. The solution involves using kinematic equations and trigonometry to determine the time of fall and the resultant velocity components.

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goj2
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Ive been trying to get this problem. I already have the answers but I want to know how to solve it.

An airplane with a speed of 77.5 m/s is climbing upward at an angle of 34.1 ° with respect to the horizontal. When the plane's altitude is 585 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.

The answers are:
(a)1040 m
(b)-61.0°

Can someone help me?
 
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what have you done so far??
 
goj2: a) work out how long the package will take to fall 585m. Then multiply the answer by the forward speed of the aircraft, worked out using trigonometry. b) is more trigonometry based on the forward speed and the vertical speed.
 

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