# Chapter 5 - Force and Motion-I - Problems: 21b

59 degrees below the horizontal

#### Work Step by Step

To find the direction of the acceleration vector we need to make a triangle with the x and y components as the two legs. To find the direction, we need to find the angle it makes with the horizontal which can be found by $tan(\theta) = \frac{y}{x}$ and therefore $\theta = arctan(-10/6) = -59^{\circ}$

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