# Chapter 3 - Kinematics in Two Dimensions; Vectors - Problems - Page 70: 31

481m

#### Work Step by Step

Let x= 0 and y = 0 directly below where the package is launched, and let upward be the positive y direction. The projectile’s initial vertical velocity is $(69.4 m/s)(sin 0 ^{\circ}) = 0$. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. It starts at a height of y = 235 m, and the final vertical position is 0 m. Use equation 2.11b to find the time of flight of 6.925 s. The horizontal distance covered is the constant horizontal velocity multiplied by the time. $$(69.4 m/s)(cos 0 ^{\circ})(6.925 s) = 481 m$$ That’s how far in advance the goods should be dropped.

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