- #1

- 35

- 0

## Homework Statement

A long jumper jumps with an angle of 23 degrees and lands 8.59m away from the jump spot. What is the magnitude of the initial velocity of the jumper?

Assumptions:

The feet land in the same place as the body (the body is to be treated as an object)

Horizontal motion is constant

**Given: (small "v" is speed, large "v" is velocity)**

x - components

a = 0

V

_{i}= vcos23

d = 8.59m

t = ?

y - components

a = 9.81m/s/s

V

_{i}= vsin23

d = ? (assumed zero at landing)

t = ?

## Homework Equations

t = d/v

at = v

_{f}- v

_{i}

d = v

_{i}+ 1/2at

^{2}

## The Attempt at a Solution

t = 8.59m / vcos23

from this point, I forgot explicitly the steps I took, but I subbed in what I knew for the initial velocities in an attempt to solve them. I ended up with:

-0.3907v

^{2}-0.3907v + 91.2 = 0

I had included units in all my procedures, and the only units left in the equation were attached to the 91.2 and became m

^{2}s

^{-2}(metres squared per seconds squared). I went on and used the quadratic formula anyway, despite the unit anomaly and came up with 14 m/s (rounded).

I am not sure if this is even remotely correct, and if I am not even on the right track, could anyone show me how to properly solve it? It seems I'm lacking a bit in the math department, though I am at the top of my physics class (top three anyway).