Posted by **sara** on Wednesday, September 22, 2010 at 8:33am.

An athlete executing a long jumper leaves the ground at a 27.0 degree angle and travels 7.89 m. What was the takeoff speed? If this speed were increased by just 5.0%, how much longer would the jump be?

- math/ physics -
**MathMate**, Wednesday, September 22, 2010 at 8:55am
Let the angle with the horizontal be φ.

and u the take-off speed (m/s)

t = airborne duration (seconds)

The vertical initial velocity, uy

= u sin(φ)

Horizontal velocity, ux

= u cos(φ)

When she lands, vertical distance=0:

0=uy t - (1/2)gt²

t=2(uy)/g

=2u sin(φ)/g

Horizontal distance (with constant velocity ux)

= ux t

= u cos(φ) * 2u sin(φ)/g

= u²sin(2φ)/g

Can you take it from here?

- math/ physics -
**sara**, Wednesday, September 22, 2010 at 9:03am
Yes. I can handle it from here. I just needed a place to start. Thanks MathMate!

- math/ physics -
**MathMate**, Wednesday, September 22, 2010 at 9:14am
You're welcome!

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