A Simple-ish Kinematics Problem

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SUMMARY

The discussion focuses on calculating the takeoff speed of a long jumper who leaves the ground at a 45-degree angle and lands 9.1 meters away. The key equations used are P_{x}(t) = V_{0}cos(theta) t and P_{y}(t) = V_{0}sin(theta) t + 1/2gt^{2}, where g is -9.81 m/s². The user initially struggles with variable substitution but ultimately resolves the confusion with assistance from other forum members. The solution involves solving two equations with two unknowns to find the initial velocity, V_{0}.

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  • Understanding of kinematic equations
  • Knowledge of projectile motion concepts
  • Familiarity with trigonometric functions
  • Basic grasp of gravitational acceleration (g = -9.81 m/s²)
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Lancelot59
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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?
 
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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}.
 
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!
 

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