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

Answer: 1.9m/s.

Explanation: 1.9m/s.

Answer 2

The horizontal range formula tells us

`x=\frac{{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=\frac{{v_i}^2\sin50^\circ}{9.8(\rm m)/(\mathrm s^2)}\implies v_i=4.4(\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(\rm m)/(\rm s)`

`v_(yi)=v_i\sin25^\circ=1.9(\rm m)/(\rm s)`