Initial velocity of projectile given angle and max height

  • Thread starter ulfy01
  • Start date
  • #1

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.

Answers and Replies

  • #2
Gold Member
Your method is correct.
At the top of the flight, there is no vertical component of velocity, so [itex] v_y = 0 [/itex] there. You are also correct that the angle is involved. Indeed, [itex] v_o\sinθ= v_{oy} [/itex] and it is this [itex] v_o [/itex] that you are solving for.
  • #3
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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