1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Projectile motion given theta and x

  1. Jul 5, 2009 #1
    1. The problem statement, all variables and given/known data

    An Olympic long jumper leaves the ground at an angle of 23o and travels through the air for a horizontal distance of 8.7m before landing. What is the takeoff speed of the jumper?

    2. Relevant equations

    8.7m = cos(23o)Vot

    y = sin(23o)t + (1/2)(-9.8m/s2)t2

    0m2/s2 = sin2(23o)Vo2 + 2(-9.8m/s2)y

    3. The attempt at a solution

    Fifteen minutes and many permutations later, I get t = .835, .782 with Vo = 11.32m/s, 12.09m/s

    The answer in the book is 11m/s.

    Can you explain the steps you would take to solve this? It literally took me 15 minutes of mathematical manipulation to isolate t. The final step to find t for me was:

    .697 = t2(3.6934 - 4.9t2)

    Thanks guys!
  2. jcsd
  3. Jul 5, 2009 #2
    y = sin(23)*V0*t + (1/2)(-9.8m/s2)t
    u missed a Vo here.
    well get t from the first equation, and substitute it in t in the second one.
    from the first equation u get
    t=8.7 / cos(23)Vo
  4. Jul 5, 2009 #3


    User Avatar
    Homework Helper

    An easier way is to solve for t first in y = sin(23)*V0*t + (1/2)(-9.8m/s2)t^2:

    0=sin(23)*V0*t + (1/2)(-9.8m/s2)t^2
    Divide out t...
  5. Jul 6, 2009 #4


    User Avatar
    Homework Helper

    [tex]x=V_o \cos(\theta_o)t[/tex]


    [tex]t = x\frac{x}{V_o \cos(\theta_o)}[/tex]

    and from

    [tex]y = V_o \sin(\theta_o)t - 0.5gt^2[/tex]

    we have that

    [tex]y = x\tan(\theta_o) - \frac{gx^2}{2V_o\cos^2(\theta_o)}[/tex]

    the parabolic equation describing the trajectory of the projectile
    Last edited: Jul 6, 2009
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook