Simple Initial Velocity Problem

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SUMMARY

The discussion centers on calculating the takeoff speed of an Olympic long jumper who leaves the ground at an angle of 18.3° and covers a horizontal distance of 8.86 m. The initial approach involved incorrectly assuming a straight diagonal path, leading to an erroneous calculation of 8.52 m/s for the takeoff speed. The correct method requires applying projectile motion equations, specifically focusing on vertical and horizontal components separately to derive two equations with two unknowns. Utilizing the projectile range equation will provide a more accurate solution.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Basic trigonometry, particularly sine and cosine functions
  • Knowledge of gravity's effect on motion (g = 9.81 m/s²)
NEXT STEPS
  • Research the "projectile range equation" for horizontal distance calculations
  • Study kinematic equations for vertical motion analysis
  • Learn how to decompose vectors into horizontal and vertical components
  • Practice solving similar problems involving angles and distances in projectile motion
USEFUL FOR

Students studying physics, educators teaching projectile motion, and anyone interested in solving real-world motion problems in sports science.

derekfuselier
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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!
 
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derekfuselier said:
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!
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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