Question

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)

Answer 1

The horizontal range formula tells us

$x=\dfrac{{v_i}^2\sin2\theta}g$

where $v_i$ is the dancer's initial speed, $\theta$ is her launch angle, and $g$ is the acceleration due to gravity. So

$1.5\,\mathrm m=\dfrac{{v_i}^2\sin50^\circ}{9.8\frac{\rm m}{\mathrm s^2}}\implies v_i=4.4\dfrac{\rm m}{\rm s}$

Her initial velocity as a vector is

$\vec v_i=v_{xi}\,\vec\imath+v_{yi}\,\vec\jmath$

where

$v_{xi}=v_i\cos25^\circ=4.0\dfrac{\rm m}{\rm s}$

$v_{yi}=v_i\sin25^\circ=1.9\dfrac{\rm m}{\rm s}$

Answer 2

Answer: 1.9m/s.

Explanation: 1.9m/s.