- #1

Peter Halsall

- 11

- 0

## Homework Statement

An athlete can cover 3.0 m for a standing long jump. Assuming that this athlete jumps at an angle of 45 degrees to the horizontal. calculate the speed at which the jumper can launch herself from a standing start.

Assume g =10 ms^-2

and no air resistance

Assume that horizontal velocity never changes

## Homework Equations

v=u+a*t

x=(u+v/2)*t

x=u*t+(a/2)*t^2

x=v*t-(a/2)*t^2

v^2=u^2+2*a*x

v= final velocity

x= displacement

a= acceleration

u= initial velocity

t= time

sin theta*initial velocity= vertical initial velocity

cos theta*initial velocity= horizontal initial velocity

## The Attempt at a Solution

In my scientific calculator I observed that cos 45 = sin 45 and thus the initial vertical and horizontal velocity must be equal to each other. From my observations, I do not think I have enough information to find this from the formula's listed. I don't really know what to do from here =/