Initial velocity of projectile given angle and max height

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SUMMARY

The discussion focuses on calculating the initial velocity of a projectile launched at an angle of 43.00 degrees, reaching a maximum height of 2974 meters. The key equation used is Vy² = V0y² - 2g(y - y0), where at the peak of the trajectory, the vertical component of velocity (Vy) is zero. The relationship between the initial velocity (V0) and its vertical component (V0y) is established through the equation V0y = V0 * sin(θ). The correct application of these principles leads to the successful calculation of the initial speed.

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ulfy01
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Homework Statement


A projectile was fired across level ground at an initial angle of 43.00 degrees above the horizontal. During its flight, it reached a maximum height of 2974 metres. What was the initial speed of this projectile?


Homework Equations



I assume that the correct equation to use is Vy2 = V0y² - 2g(y - y0).

The Attempt at a Solution



To be honest I'm not even sure how to go about it. Given the equation, I would rearrange for V0y² however that doesn't seem to work at all. I'm not even sure what the value of Vy has to be. I also suspect the angle comes into play, but I'm not sure what it can tell me. Any pointers greatly appreciated.
 
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Your method is correct.
At the top of the flight, there is no vertical component of velocity, so v_y = 0 there. You are also correct that the angle is involved. Indeed, v_o\sinθ= v_{oy} and it is this v_o that you are solving for.
 
You're a life savior. It's this vo sinθ= voy that I wasn't able to get to. I got the answer right. Thank you so much!
 

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