Conservation of Energy + Projectile Motion

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Homework Help Overview

The discussion revolves around a problem involving the conservation of energy and projectile motion. The original poster describes a scenario where a ball rolls down a ramp and launches at a 15-degree angle, landing 100 cm away in a hole that is 0.05 cm deep. The goal is to determine the height from which the ball must be rolled down the ramp to achieve this landing distance.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to use conservation of energy to calculate the launch velocity and then apply projectile motion equations to find the required height. Some participants question specific components of the calculations, such as the interpretation of the distance and the mixing of units.

Discussion Status

Participants are exploring the problem with some guidance provided regarding the overall approach. There are indications of confusion about specific calculations and unit consistency, but no consensus has been reached on the correctness of the original poster's method.

Contextual Notes

There is a mention of potential errors in algebra and the need for clarity in the formulas used. The original poster notes that this problem is part of a lab, which may imply additional constraints or context not fully detailed in the discussion.

andy343
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Homework Statement


So I have a ball rolling down a ramp structure from rest and up a curve,to launch out at 15 degrees. The ball lands 100cm away in a hole 0.05cm deep.
The height in which it comes out of the ramp is 0.05cm high
angle at which ball launches out = 15 degrees above horiontal
dx=100cm
dy=-0.05+-0.05= -0.1cm
*also the lowest point of the ramp is on the ground

find the height in which the ball needs to be rolled down the ramp in order to land 100cm away and into a hole 0.05cm deep.

Homework Equations


Conservation of energy
projectile motion equations

The Attempt at a Solution


Use conservation of energy equation to find velocity and than input that into dy=vyit+1/2ayt^2
Ei=Ef
Ei=KEt+KEr
mgh=1/2mv^2+1/2Iw^2
I=2/5mr^2
w=v/r
mgh=7/10mv^2
solve for v, v=sqrt(10/7gh)---= launching velocity
Use formula dy=vyit+1/2ayt^2
t=dx/vx
vx=cos(15)vi
vyi=sin(15)vi
-0.1=sin15vi(100/cos15vi)+1/2(-9.81)(1/(cos15)(vi))^2
vi cancels vi=launching velocity = sqrt(10/7gh)
-0.1=sin15(100/cos15)+1/2(-9.81)(100^2/(cos15)^2(10/7(9.81)h)
9.81 cancels
-0.1=sin15(100/cos15)+1/2(-1)(100^2/(cos15)^2(10/7h)
solve for h
h=-70.9cm
Umm, I don't get why my height becomes negative..so I think I did something wrong but
I don't know where. ---This is actually from my lab and not a problem so I might be missing something.
 
Last edited:
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I'm slightly confused about what the 100/cos15 is. Is that your time?
 
Um oh sorry about the confusion...I accidently inputed some wrong numbers, give me a second
and yes that is my time
 
Last edited:
Where does "100" come from?
 
So it lands in a hole 100cm* away in 0.05cm* deep hole.
 
Last edited:
You can't mix meters and centimeters in one formula. "9.81" is in meters.
 
lol, okay I think I know why my calculations are wrong now...my bad
TYVM
One more thing, so am I approaching the calculation/problem correctly? by using conservation of energy to find launch velocity and than input that into the dy=vyit+1/2ayt^2?
 
The overall approach seems OK. There might be errors in algebra - the formulas are very hard to read.
 

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