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?

Answer 1

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?

Answer 2

Yes. I can handle it from here. I just needed a place to start. Thanks MathMate!

Answer 3

You're welcome!