Finding initial speed given two angles and a distance

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Homework Help Overview

The problem involves a child running down an inclined plane and jumping at an angle, with the goal of determining the initial speed required to land a specific distance down the hill. The subject area includes concepts from projectile motion and trigonometry related to inclined planes.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using projectile motion equations to relate vertical and horizontal components of motion. There are attempts to express time and initial speed in terms of the angles and distance involved. Questions arise regarding the specific equations to use and how to isolate the initial speed from the derived equations.

Discussion Status

The discussion is active, with participants providing suggestions for setting up equations and expressing variables. Some participants have expressed uncertainty about isolating the initial speed and have encountered issues with negative values under square roots, indicating a need for further clarification or alternative approaches.

Contextual Notes

There are indications of confusion regarding the setup of the equations and the assumptions made about the initial and final positions in the context of the problem. The specific angles and distances involved are critical to the discussion.

heyhowsitgoin
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Homework Statement



A child runs down a 11 degree hill and then suddenly jumps upward at a 16degree angle above the horizontal and lands 1.2 m down the hill as measured along the hill

What was the child's initial speed?

Homework Equations



Projectile motion

x = Vx*t

y = yo + Vyo*t - 1/2gt^2

The Attempt at a Solution



tried solving for time t from the vertical motion equation y = yo+ Vyo*t - 1/2gt^2 and plug it in the
x = vx*t..but it doesn't work
 
Last edited:
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Hi heyhowsitgoin
Welcome to PF.
If you draw the figure of inclined plane and the landing of the child, you can see that
y = 1.2*sin11.
t = x/vx = 1.2*cos11/vo*cos16
vyo = vo*sin16.
Substitute these values in the equation and solve for vo.
 
Hi thanks for the reply. Which equation are you referring to solve for vo?
 
heyhowsitgoin said:
Hi thanks for the reply. Which equation are you referring to solve for vo?

y = yo+ Vyo*t - 1/2gt^2
 
hrmm ok

so i end up with 0.229 = 0 + Vo*sin16 (1.2*cos11 / Vo*cos16) - 1/2(9.8)(1.2*cos11/Vo*cos16)^2

is this right? not sure how to solve for Vo at this point.
 
It is right. Now simplify everything. And solve for Vo.
 
Hello,

I was looking at this question and was not able to isolate Vo in the final equation. I ended up with a negative number under a square root...
Any suggestions on how to correctly isolate Vo?
 
heyhowsitgoin said:
hrmm ok

so i end up with 0.229 = 0 + Vo*sin16 (1.2*cos11 / Vo*cos16) - 1/2(9.8)(1.2*cos11/Vo*cos16)^2

is this right? not sure how to solve for Vo at this point.
In the equation final position y is zero, and initial position yo is 0.229
 

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