# Question concerning initial velocity

## 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
Homework Helper
Are you computing

distance=V0sin2θ/g

OR

Distance = V02sin2θ/g?

The second one is the one should use.

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

Thank you so much!

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)