Motion in Two or Three Dimension : Projectile motion

AI Thread Summary
A projectile is launched at a 60° angle, and the change in its velocity is given as Δv = -20 y^ m/s. The discussion revolves around calculating the initial and final velocities, with participants attempting to apply kinematic equations. There is confusion regarding the final vertical position (Yf - Yo) since the projectile is launched and lands on the same level ground. Some participants question the accuracy of the problem statement and the algebra used in the calculations. The conversation highlights the importance of clear problem wording and correct application of physics equations in solving projectile motion problems.
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Homework Statement



A projectile is launched at a 60° angle above the horizontal on level ground. The change in its velocity between launch and just before landing is found to be Δv = vlanding _ vlaunch = -20 y^ m/s .
What is the initial velocity of the Projectile ?
What is its final velocity just before landing ?

Homework Equations



Δv = vlanding _ vlaunch = -20 y^
m/s .
3.The attempt at a solution
I considered :
Vox - Vfx = 0 and Voy - Vfy = -20
knowing that V²fy - V²oy= -2g (Yf-Yo) But I couldn't go further

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
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physicos said:
Vox + Vfx = 0 and Voy + Vfy = -20
"+"?
knowing that V²fy - V²oy= -2g (Yf-Yo)

What is the value of Yf-Yo?
 
Yes , just saw it ! I edited the post !
I have no idea about Yf-Yo ,I couldn't guess it from the problem statement !
 
physicos said:
Yes , just saw it ! I edited the post !
I have no idea about Yf-Yo ,I couldn't guess it from the problem statement !
It is launched from, and lands on, the same level ground.
 
so V landing = 0 ?? Because we know that v = sqrt(Vo²-2g(Y-Yo))
 
physicos said:
so V landing = 0 ?? Because we know that v = sqrt(Vo²-2g(Y-Yo))
No, you should check your algebra on that equation. But regardless, there are serious flaws in the problem statement. Did you word it correctly?
 
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