Kinematics - can't find the initial velocity according to the image

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The discussion focuses on solving a kinematics problem involving projectile motion, where the initial velocity needs to be determined from given parameters. Users are advised to write separate equations for vertical and horizontal motion, noting that horizontal velocity remains constant if air resistance is neglected. The correct application of trigonometric functions, specifically sine and cosine, is crucial for resolving the initial velocity into its components. Additionally, a suggested approach involves treating the problem in two parts: first calculating the initial speed for a projectile fired at an angle, and then determining the horizontal distance from a cliff. This method clarifies the relationship between the two scenarios and aids in solving for the unknowns.
Physics_learner
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Homework Statement
Kinematics - finding the initial velocity and final x position
Relevant Equations
x(t)=x0+v0*(t2-t1)+0.5a*(t2-2a)^2
v(t)=v0+a*(t2-t1)
I tried to write the data I understood from the image:
y0=160m
yf=0
x0=0
x1=192m
I tried to express the total change in time using the position over time equation on the Y direction:
y(t)=y0+v0y*(t2-t1)-0.5a(t2-t1)^2
but then I stuck with 2 variables and didn't know what to do

any help?
 

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Physics_learner said:
Homework Statement:: Kinematics - finding the initial velocity and final x position
Relevant Equations:: x(t)=x0+v0*(t2-t1)+0.5a*(t2-2a)^2
Welcome to PF.

It looks like you are on the right track, except you seem to have a typo in your last relevant equation (in the ##t^2## term). Can you fix that?

And you are taking the right approach, writing one equation for the motion in the y direction, and now you should add the 2nd equation which is for the motion in the x direction. Hint: the x-direction velocity does not slow down or change, if you neglect air resistance.

Then for your two equations, you use ## sin \theta ## and ## cos \theta ## appropriately to give you the components of the initial velocity in the two directions (y and x), and work on solving the two unknowns using the two equations. Does that help?
 
A different approach, which is perhaps conceptually easier to implement, is to split the problem in two parts. Note that when the projectile returns to the same height at the corner of the step, its speed is the same but the vertical component has changed direction and is down instead of up.

To find the initial speed ##v_0##, solve the problem "A projectile is fired at 37° above the horizontal and travels 192 m horizontally before it returns to the same height. Find the initial speed ##v_0##."

To find ##x## in the figure, solve the problem "A projectile is fired from the edge of a 160 m cliff at 37° below the horizontal. When it lands, it is horizontal distance ##x## from the edge. The initial speed is the same as in the previous problem. Find ##x##." (Don't forget that the final horizontal position is 192 + ##x~##meters.)
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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