Question concerning initial velocity

[Casimi]

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.

 

[rock.freak667]

Are you computing

distance=V₀sin²θ/g

OR

Distance = V₀²sin2θ/g?

The second one is the one should use.

 

[Casimi]

It seems as if I derived the wrong formula via a mistake in my algebra.

Thank you so much!

 

[The legend]

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 answerEDIT: rock.freak answered first, i guess...

 

[Casimi]

It seems like I should take more time to organize my thoughts and perform my algebra correctly. Simple mistakes are always my downfall!