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Simple Initial Velocity Problem

  1. Feb 22, 2007 #1
    An Olympic long jumper leaves the ground at an angle of 18.3 ° and travels through the air for a horizontal distance of 8.86 m before landing. What is the takeoff speed of the jumper?

    I'm completely dumbfounded on this one. What I have done is cut the distance in half to create a triangle and then solve all sides in order to find the y. Once I have the way I use the formula ( t = sq rt of 2y/g ) to get the time for the object to fall half way. I multiplied by two to get total time of flight (or what I thought was total time of flight.) Once I have the alleged total time of flight I divide 8.86 by it, giving me the velocity in the x direction. Using cos(18.3) = 8.09 m/s divided by the hypotenuse, or the initial velocity. Solving for the initial velocity left me with 8.52 m/s, which is wrong.

    Can anyone offer any help? I can scan my work if needed, I know it's hard to read and think about math like this. Thanks in advance!
     
  2. jcsd
  3. Feb 23, 2007 #2

    PhanthomJay

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    Your error is in assuming that the jumper moves in a staight diagonal line to a max height at the apex of the triangle you have described. Since gravity is always acting downward and causing the jumper to continually 'fall' as he jumps, his motion will be parabolic in accordance with projectile motion equations of motion. You should look first in the vertical direction using the appropriate kinematic equation that relates vertical distance with initial vertical speed and time, noting that y = 0 as the jumper hits the ground. Then your second equation comes from looking in the horizontal direction as you have noted. You get 2 equations with 2 unknowns, which can be solved with a bit of trig manipulation. I suggest you do a google search on "projectile range equation" to see an example of a similar problem.
     
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