Posted by sara on .
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,
Let the angle with the horizontal be φ.
and u the takeoff 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,
Yes. I can handle it from here. I just needed a place to start. Thanks MathMate!

math/ physics 
MathMate,
You're welcome!