Question concerning initial velocity

  • Thread starter Casimi
  • Start date
  • #1
Casimi
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Homework Statement



An athlete executing a long jump leaves the ground at a 31.6° angle and travels 7.78 m. What was the take-off speed?

I have tried to solve this question but somehow keep arriving at the wrong answer. The formula that I derived is :

distance=V(initial)*sin^2(theta)/g

Where am I going wrong here? Any help would be appreciated.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,223
31
Are you computing

distance=V0sin2θ/g

OR

Distance = V02sin2θ/g?

The second one is the one should use.
 
  • #3
Casimi
11
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It seems as if I derived the wrong formula via a mistake in my algebra.

Thank you so much!
 
  • #4
The legend
426
0
i get a different formula for it.

First take the vertical component, and find the total time taken. Now take the horizontal component of the velocity and multiply by time to find the expression for distance(range)

Substitute and get your answer


EDIT: rock.freak answered first, i guess...
 
  • #5
Casimi
11
0
It seems like I should take more time to organize my thoughts and perform my algebra correctly. Simple mistakes are always my downfall!
 

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