Ask Question
23 August, 12:33

An athlete performing a long jump leaves the ground at a 27.0 degree angle and lands 7.80m away. What was the takeoff speed?

+3
Answers (1)
  1. 23 August, 16:19
    0
    =9.72 m/s

    Explanation:

    From the Newton's laws of motion;

    x=2 (v²cos∅sin∅) / g

    Using geometry we see that 2 cos∅sin∅ = sin 2∅

    Therefore, x = (v²sin 2∅) g, where v is the take off speed x the range and ∅ the launch angle.

    Making v the subject of the formula we obtain the following equation.

    v=√{xg / (sin 2∅) }

    x=7.80

    ∅=27.0

    v=√{7.8*9.8/sin (27*2) }

    v=√94.485

    v=9.72 m/s
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “An athlete performing a long jump leaves the ground at a 27.0 degree angle and lands 7.80m away. What was the takeoff speed? ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers