50 POINTS! A Dancer leaps across the stage in a beautiful grand jete. The ballerina leaps at an angle of 25° and is able to jump a horizontal distance of 1.5m. What is her initial x and initial y velocity? (hint: find Vi first, then Vxi and Vyi)
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The horizontal range formula tells us
[tex]x=\dfrac{{v_i}^2\sin2\theta}g[/tex]
where [tex]v_i[/tex] is the dancer's initial speed, [tex]\theta[/tex] is her launch angle, and [tex]g[/tex] is the acceleration due to gravity. So
[tex]1.5\,\mathrm m=\dfrac{{v_i}^2\sin50^\circ}{9.8\frac{\rm m}{\mathrm s^2}}\implies v_i=4.4\dfrac{\rm m}{\rm s}[/tex]
Her initial velocity as a vector is
[tex]\vec v_i=v_{xi}\,\vec\imath+v_{yi}\,\vec\jmath[/tex]
where
[tex]v_{xi}=v_i\cos25^\circ=4.0\dfrac{\rm m}{\rm s}[/tex]
[tex]v_{yi}=v_i\sin25^\circ=1.9\dfrac{\rm m}{\rm s}[/tex]