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A long jumper leaves the ground at 4545^{\circ} above the horizontal and lands 8.0 m8.0 \mathrm{~m} away. What is her "takeoff" speed v0v_0?

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(a)(a) \\ Use the range formula (level horizontal range), derived in Example 3-10\ to find her takeoff speedv0v_{0}.\\$R=\displaystyle $$\frac{v_{0}^{2}\sin 2\theta_{0}}{g}$

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