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I have the question "A particle is projected so that the horizontal and vertical components of its initial velocity are U and V respectively. Show that the maximum height of the particle is:

V^2 / 2g "

Here is my attempt however I do not get the correct result what have I done wrong ?

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Dan
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    Can you explain the first line? Where does that equation come from? – Matthew Leingang Dec 05 '17 at 00:13
  • It's the equation used to find the maximum height S as the equation consists of what is needed to find this as it involves V, U and the height S – Dan Dec 05 '17 at 00:15
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    The equation $v^2=u^2+2as$ refers to one-dimensional motion in which $u$ is the initial velocity and $v$ is the final velocity. This is not the case in your problem. Also, I suspect that $u=v\sin\alpha$ comes from a different interpretation again. An excellent illustration of the danger of memorising formulae without actually understanding what they mean. – David Dec 05 '17 at 00:15
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    How is $U$ related to the height? It's the horizontal component of the initial velocity. – Matthew Leingang Dec 05 '17 at 00:16
  • Yes that's true I've used the wrong equation thanks :) – Dan Dec 05 '17 at 00:17
  • But I'm not sure which equation to use ? Because this is the only equation which consists of S, U , V and a – Dan Dec 05 '17 at 00:21
  • It's actually the right equation I found the solution online :) thanks anyway – Dan Dec 05 '17 at 00:26

1 Answers1

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At maximum height, the vertical velocity is zero. We also know that in simple kinematic situations with constant acceleration:

$$V_f^2 = V_0^2 + 2ax$$

And for our situation, $V_f = 0$, $V_0 = v$, $a = -g$, and $x = h$.

So we have

$$0^2 = v^2 - 2gh$$ Solve for $h$ to get

$$h = \frac{v^2}{2g}$$

actinidia
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