Mechanics: calculating launch angle of projectile

[haruspex]

Thank you so much for all the explanations. I have only one more follow up question. In post #16 I used one of the kinematic equations to find the "initial vertical velocity":

$y-y_{0}=v_{y}t+\frac{1}{2}at^{2}$
$9-0=v_{y}(4.5)+\frac{1}{2}(-9.81)(4.5)^{2}$
$\therefore v_{y}\approx\ 24 \text{m/s.}$

Is it possible to use this same value to work out the maximum height $h$ of the projectile (at point B of the diagram)?

I mean, there is another kinematic equation that could be used (given that the vertical velocity at the apex is 0):

$v_{y(\text{final})}^{2}=v_{y(\text{initial})}^{2}+2ah$
$0=24^{2}+2(-9.8)h \implies h \approx 29\ \text{m}$

So, was our previously calculated value for $v_{y}$ appropriate to be substituted as $v_{y(\text{initial})}$ in the second equation for finding $h$?

Yes, but finding one numerical value then plugging that into a formula to find another can lead to accumulation of errors. Keep a digit or two of extra precision in the first calculation.
Alternatively, it is often better to keep the first result in algebraic form and continue that way to the second result, only plugging in numbers at the final step. That might be awkward here, though.

 

[rudransh verma]

So, was our previously calculated value for vy appropriate to be substituted as vy(initial) in the second equation for finding h?

Of course. Vy will not change. It’s same for all instants in this situation.

 

[neilparker62]

I mean, there is another kinematic equation that could be used (given that the vertical velocity at the apex is 0):

$v_{y(\text{final})}^{2}=v_{y(\text{initial})}^{2}+2ah$
$0=24^{2}+2(-9.8)h \implies h \approx 29\ \text{m}$

So, was our previously calculated value for $v_{y}$ appropriate to be substituted as $v_{y(\text{initial})}$ in the second equation for finding $h$?

Conservation of mechanical energy: $$mgh_{max}=½m{v_{yi}}^2$$

 

[kuruman]

$\dots$
I mean, there is another kinematic equation that could be used (given that the vertical velocity at the apex is 0) $\dots$

You can invent your own equation in terms of the given quantities. Knowing the time of flight makes the task easy. You know that the maximum height occurs at half the time of flight and that the vertical velocity is zero when that happens. Then you can write two equations expressing these ideas:
$$h_{\text{max}}=v_{\text{0y}}\frac{t_{\!f}}{2}-\frac{1}{2}g\left(\frac{t_{\!f}}{2}\right)^2~;~~0=v_{\text{0y}}^2-2gh_{\text{max}}.$$The task before you is to use the second equation to substitute $v_{\text{0y}}$ into the first equation and solve the ensuing quadratic.

On edit: This is bad advice. See post #36.

 

[PeroK]

Is it possible to use this same value to work out the maximum height $h$ of the projectile (at point B of the diagram)?

If you know that it was at height $y = 9m$ at time $t = 4.5s$, then you have:
$$y = v_{0y}t - \frac 1 2 g t^2 \ \Rightarrow \ v_{0y} = \frac 1 {t}(y + \frac 1 2 g t^2)$$And, the maximum height $h$ is given by the equation $$2gh = v_{0y}^2 \ \Rightarrow \ h = \frac{v_{0y}^2}{2g}$$ And, as mentioned above, you can either combine those equations or solve for $v_{0y}$ first.

 

[PeroK]

You can invent your own equation in terms of the given quantities. Knowing the time of flight makes the task easy. You know that the maximum height occurs at half the time of flight and that the vertical velocity is zero when that happens. Then you can write two equations expressing these ideas:
$$h_{\text{max}}=v_{\text{0y}}\frac{t_{\!f}}{2}-\frac{1}{2}g\left(\frac{t_{\!f}}{2}\right)^2~;~~0=v_{\text{0y}}^2-2gh_{\text{max}}.$$The task before you is to use the second equation to substitute $v_{\text{0y}}$ into the first equation and solve the ensuing quadratic.

You do need to be careful if $t_f$ is not when it returns to the ground.

 

[kuruman]

You do need to be careful if $t_f$ is not when it returns to the ground.

Oops! I lost myself somewhere.