0
$\begingroup$

I am learning about projectiles in my physics course and was recently introduced to this equation: $$\Delta y_{max} = \frac{v_0^2\sin^2(\theta)}{2y}$$ I am having a hard time making sense of it. How does one utilize it? When are you supposed to use it? How do you put it in the calculator? I ask because I don't know how to input sin^2, and I only see Sin and Sin inverse within the calculator.

Thank you

$\endgroup$

2 Answers 2

1
$\begingroup$

The equation is used to find the maximum height of an object undergoing projectile motion. Note it depends only on the initial vertical velocity, angle of launch and gravitational acceleration. Also $\sin^2(x) = (\sin x)^2$.

This question is probably more appropriate for physics stack exchange.

$\endgroup$
1
  • $\begingroup$ You didn't say/write too much but it was just what I needed. Thank you. $\endgroup$
    – יהודה
    Oct 8, 2021 at 4:02
1
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$


\begin{align} 0 & = \totald{\bracks{v_{0}\sin\pars{\theta}t - gt^{2}/2}}{t} \implies t = {v_{0}\sin\pars{\theta} \over g} \\[5mm] & v_{0}\sin\pars{\theta}t - {1 \over 2}\,gt^{2} = v_{0}\sin\pars{\theta}\, {v_{0}\sin\pars{\theta} \over g} - {1 \over 2}\,g\bracks{v_{0}\sin\pars{\theta} \over g}^{2} = \bbx{\color{#44f}{v_{0}^{2}\sin\pars{\theta} \over 2g}} \\ & \end{align}
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.