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March 2, 2015

March 2, 2015

Posted by **teamjakeward** on Wednesday, February 11, 2009 at 8:30pm.

A track star in the long jump goes into the jump at 12 m/s and launches herself at 20.0degrees above the horizontal. What is the magnitude of her horizontal displacement?

(assume no air resistance and that

Ay= -g = -9.81 m/s^2)

thanks :)

- physics -
**drwls**, Wednesday, February 11, 2009 at 8:52pmHer vertical component of velocity at takeoff is

Vyo = 12 sin 20 = 4.10 m/s

and her horizontal component is

Vx = 12 cos 20 = 11.28 m/s.

The horizontal component remains constant during the jump.

Use the vertical velocity component to determine how long she is in the air before hitting the ground. Call that time T

Vo = g T/2

T = 2 Vo^2/g

The length of the jump is Vx*T

= 2 (Vo^2/g) sin 20 cos 20

= (Vo^2/g) sin 40

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