# Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 49

(a) $v_0 = 12.3~m/s$ (b) The maximum height above the ground is 0.90 meters.

#### Work Step by Step

(a) $x = \frac{v_0^2~sin(2\theta)}{g}$ $v_0^2 = \frac{gx}{sin(2\theta)}$ $v_0 = \sqrt{\frac{gx}{sin(2\theta)}}$ $v_0 = \sqrt{\frac{(9.80~m/s^2)(10~m)}{sin(2\times 20^{\circ})}}$ $v_0 = 12.3~m/s$ (b) $2ah=v_y^2-v_{0y}^2$ $2(-g)h=v_y^2-v_{0y}^2$ $h=\frac{v_{0y}^2-0}{2g}$ $h=\frac{[v_0~sin(\theta)]^2-0}{2g}$ $h=\frac{[v_0~sin(\theta)]^2}{2g}$ $h=\frac{[(12.3~m/s)~sin(20^{\circ})]^2}{(2)(9.80~m/s^2)}$ $h = 0.90~m$ The maximum height above the ground is 0.90 meters.

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