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A quadratic of the form

$$h(t)=gt^2+v_0t+h_0$$

where $t$ is time in seconds, $g$ is acceleration due to gravity, and $h(t)$ is height in meters - can be used to approximate the theoretical (no wind resistance etc) trajectory of a projectile. If $h_0=0$ and the ground is level, the initial velocity is $v_0$, and the firing angle is $\theta = \arcsin{(v_0)}$. Accordingly, increasing $v_0$ rapidly increases $\theta$.

Here is my question:

At the level of a high school geometry student, how can I change my $v_0$ while leaving $\theta$ constant. In other words, how can I modify my $h(t)$ equation so that $\theta$ is fixed but $v_0$ can be modified? Assume understanding of basic trig ratios, but very little understanding of trig functions. For instance, how could I obtain a quadratic model with a $v_0$ of 40 m/s and an initial firing angle of $\frac{\pi}{4} = 45$?

The goal is to allow students to find a particular quadratic model (that accounts for gravity correctly but not other factors) that can be used to launch a projectile to hit a target.

Posting solutions of any kind is appreciated; posting a solution in high school level math is greatly appreciated.

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    $\begingroup$ $\theta = \arcsin(v_{0,y}/v_{0})$. The argument of arsin should be dimensionless. $\endgroup$ – Tucker Jun 12 '17 at 4:37
  • $\begingroup$ That is not the trajectory of a projectile, it's only its vertical motion. In addition, you should write ${1\over2}gt^2$. And $v_0$ is vertical velocity at $t=0$, so your statement $\theta=\arcsin(v_0)$ is meaningless. $\endgroup$ – Aretino Jun 12 '17 at 6:40
  • $\begingroup$ Thanks for the good feedback. Since this question requires fatal overhaul, I've rewritten the question more correctly and more clearly here. $\endgroup$ – Zediiiii Jun 13 '17 at 13:09
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You want $\boxed{h(t)=gt^2+v_0\sin(\theta)t+h_0}$.

This is because the vertical component of the $v_0$ arrow pointed in the $\theta$ direction has height $v_0\sin(\theta)$.

By the way, in this problem $g=-4.9$, because the acceleration due to gravity is $-9.81$ m/$s^2$.

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  • $\begingroup$ Since I will be deleting this poorly-worded question, you might like to post here as well. Thanks. $\endgroup$ – Zediiiii Jun 13 '17 at 13:13

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