Monday
March 27, 2017

Post a New Question

Posted by 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 - ,

    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 - ,

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

  • math/ physics - ,

    You're welcome!

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question