Finding the max height of a ball launched as a projectile using work-energy

Question

A ball is launched as a projectile with initial speed $v$ at an angle $\theta$ above the horizontal. Using conservation of energy, find the maximum height $h_\text{max}$ of the ball's flight. Express your answer in terms of $v$, $g$, and $\theta$.

I am doing:

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

Is that correct, or is there a much easier way...?

Answer 1

Under the constraints of the problem, then yes, what you're doing is correct.

If you weren't required to use conservation of energy, then it would probably be easier to calculate the vertical component of the initial velocity and use 1D kinematics.

Answer 2

I am not sure what David is pointing at in his second paragraph, but you could also consider only the vertical component of the velocity and ignore the horizontal one. The initial vertical velocity is:

$$ v_{vert} = v \sin(\theta) $$

Using

$$ 0.5 m v^2 = m g h $$

this leads to

$$ \begin{align*} 0.5 m (v \sin(\theta))^2 = m g h_{max}\\ \frac{0.5 (v \sin(\theta))^2}{g} = h_{max}\\ h_{max} = \frac{v^2 \sin^2(\theta)}{2 g } \end{align*} $$

Answer 3

The energy supplied to a projectile be devided into two parts one along vertical velocity and other along vx since vx is constant proving the continuous velocity along x axis hence by virtue of motion the projectile has k.e through ought the journey though differing in amount