Athlete Long-Jump Question, Don't Know Where I've Gone Wrong

  • Thread starter lando45
  • Start date
  • #1
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An athlete executing a long jump leaves the ground at a 35° angle and travels 5.80 m. What was her takeoff speed?

I was set this question and have spent about an hour trying to solve it but just can't come up with the correct answer. I found this formula:

Angular Velocity w = 2Π / T
35 = (2 x 3.14) / T
T = 0.179
V = D / T
V = 5.80 / 0.179
V = 32.4


But this is wrong...is there another formula I should be using?
 

Answers and Replies

  • #2
44
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You need to find the vertical componant of the velocity, by using trigonometry.
Then you use suvat to find the time when distance is zero again (i.e. when the athlete hits the ground).

Now you know the time when he hits the ground, and using trigonometry, you can find the horizontal componant of the velocity. Then find how far they have travelled at the time when they hit the ground again
 
  • #3
siddharth
Homework Helper
Gold Member
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lando45 said:
An athlete executing a long jump leaves the ground at a 35° angle and travels 5.80 m. What was her takeoff speed?
I was set this question and have spent about an hour trying to solve it but just can't come up with the correct answer. I found this formula:
Angular Velocity w = 2Π / T
35 = (2 x 3.14) / T
T = 0.179
V = D / T
V = 5.80 / 0.179
V = 32.4

But this is wrong...is there another formula I should be using?
In the question, it is given that the angle at which the athlete leaves the ground is 35°.
This is entirely different from angular velocity.
Also, don't try to solve problems by blindly applying formulas.

For example, in this question, what happens to the athlete during the jump and why? Can you find what forces are acting on him during the jump along the horizontal and vertical directions?
 
  • #4
daniel_i_l
Gold Member
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Remember that the time in the air only has to do with the vertical componant, and the distance only has to do with the horizontal one.
 
  • #5
335
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Remember the formula for the range of a projectile. thus you will get the horizontalcomponent aswell as the takeoff velocity since the angle the initial velocity vector makes with the surface is already given.
R = v^2sin(2theta)/g
 

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