# Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 54

The initial speed of the arrow is 106 m/s

#### Work Step by Step

When the arrow hits the ground: $t = \frac{v_y}{g}$ When the arrow hits the ground: $\frac{v_y}{v_x} = tan(3.0^{\circ})$ $v_y = v_x~tan(3.0^{\circ})$ We can find the initial speed of the arrow $v_x$. $v_x~t = x$ $\frac{v_x~v_y}{g} = x$ $\frac{v_x^2~tan(3.0^{\circ})}{g} = x$ $v_x^2 = \frac{gx}{tan(3.0^{\circ})}$ $v_x = \sqrt{\frac{gx}{tan(3.0^{\circ})}}$ $v_x = \sqrt{\frac{(9.80~m/s^2)(60~m)}{tan(3.0^{\circ})}}$ $v_x = 106~m/s$ The initial speed of the arrow is 106 m/s

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