# Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 13

The plane should drop the package 678 meters short of the target.

#### Work Step by Step

We can find the time for the package to drop 100 meters. $y = \frac{1}{2}gt^2$ $t = \sqrt{\frac{2y}{g}}$ $t = \sqrt{\frac{(2)(100~m)}{9.80~m/s^2}}$ $t = 4.52~s$ The we can find the horizontal distance that the package will travel in 4.52 seconds. $x = v_x~t = (150~m/s)(4.52~s)$ $x = 678~m$ The plane should drop the package 678 meters short of the target.

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