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I'm having some difficulty understanding how to find the maximum height using conservation of energy.

This is the picture I'm looking at:

enter image description here

and this is how you find it: $$\begin{align*} \frac{1}{2}mv^2 &= mgh_\text{max} + \frac{1}{2}m(v\cos\theta)^2\\ v^2 &= 2gh_\text{max} + (v\cos\theta)^2 \\ h_\text{max} &= \bigl(v^2 - (v\cos\theta)^2\bigr)/2g \\ h_\text{max} &= v^2\bigl(1 - (\cos\theta)^2\bigr)/2g \\ h_\text{max} &= \frac{v^2\sin^2\theta}{2g} \end{align*}$$

However, I am confused about a few things. I know that all those equations stem from using $K_{i} + U_{i} = K_{f} + U_{f}$. The initial potential energy is 0 because it just started moving, correct? How come we needed to use the x-component of the kinetic energy to use $K_{f}$ (I assume that's where the cos came from) and not for $K_{i}$, where it's just $1/2mv^2$. I don't understand the importance of it?

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4 Answers 4

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The initial potential energy is zero because the ball starts off at essentially ground level, and potential energy is being defined as being zero at ground level.

The initial velocity is a vector of magnitude v that points up at an angle $\theta$ from the ground. The components of that initial velocity are $v_x(0)=v \cos\theta$ in the horizontal direction, and $v_y(0)=v \sin\theta$ in the vertical direction.

$v_y(t)$ changes with time due to gravity, with $v_y(t_{apex})=0$ when the ball is at its apex.

$v_x(t)$ doesn't change with time during the ball's path, because there is no horizontal force on the ball. Since at the ball's apex, $v_y(t_{apex})=0$ and $v_x$ is still given by $v_x(t_{apex})=v \cos \theta$, the ball's speed at the apex is $v \cos\theta$, which is why that speed is used for the ball's speed in the expression for the kinetic energy of the ball at its apex.

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There is no force on the x direction, so the acceleration is zero and the x-component velocity is constant which is know in the initial condition.

Plus the conservation of energy at the beginning and at the highest point, you will get that equation

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  • $\begingroup$ why doesn't the y-component velocity seem to matter? @luming $\endgroup$ Commented Oct 5, 2014 at 4:49
  • $\begingroup$ @FrostyStraw The kinetic energy is decreased because y-component velocity is decreased, and the height is increased. You can also calculate maximum height using $v_y$ if you like, for the height increased is due to $v_y$. $\endgroup$ Commented Oct 5, 2014 at 4:53
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Let's take a closer look at the equation: $$\frac{mv^2}{2} = mgh_\text{max} + \frac{m(v\cos\theta)^2}{2}$$ The term on the left is the initial kinetic energy of the cannonball as it leaves the cannon. This is equal to the horizontal kinetic energy plus the vertical kinetic or potential energy. At the maximum height, there is no vertical kinetic energy (since there is no vertical velocity), so all the energy is potential energy.

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P.E at certain height doesn't depend on path from where and how the projectile arrived there but it depends upon the higher postion wrt ground. At max height p.e is max so k.e will b zero to conserve E.

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