# A Simple-ish Kinematics Problem

1. Oct 4, 2009

### Lancelot59

A long jumper leaves the ground at 45o above the horizontal and lands 9.1m away. What is her "takeoff" speed?

Well I have

Dx=VxT, but I'm missing some variables, and substitution is failing me. How can I solve this?

2. Oct 4, 2009

### Vykan12

P_{x}(t) = V_{0}cos(theta) t

You can use that to relate initial velocity to time.

We know at least that gravity is acting against the jumper in the downward direction. Assuming there isn't air drag or other external forces, we have:

P_{y}(t) = V_{0}sin(theta) t + 1/2gt^{2}

P_{y}(t) will be 0 when the person lands, and g is obviously -9.81. Thus we have two equations with 2 unknowns, so we can solve for V_{0}.

3. Oct 4, 2009

### Lancelot59

I'm not really getting how this comes together.

EDIT: Nevermind, my confusion was over some of my own silliness. I get it now. Thanks for your help!