Question Number 215032 by Tawa11 last updated on 26/Dec/24
Commented by Tawa11 last updated on 26/Dec/24
$$\mathrm{I}\:\mathrm{got}\:\:\mathrm{10}.\mathrm{3}\:\mathrm{m}/\mathrm{s} \\ $$
Answered by mr W last updated on 26/Dec/24
$${U}^{\mathrm{2}} −{u}^{\mathrm{2}} =\mathrm{2}{gh} \\ $$$$\mathrm{tan}\:\theta=\frac{\sqrt{{U}^{\mathrm{2}} −{u}^{\mathrm{2}} }}{{u}}=\frac{\mathrm{2}{h}}{{d}} \\ $$$$\frac{\sqrt{\mathrm{2}{gh}}}{{u}}=\frac{\mathrm{2}{h}}{{d}} \\ $$$$\Rightarrow{u}={d}\sqrt{\frac{{g}}{\mathrm{2}{h}}}=\mathrm{8}\sqrt{\frac{\mathrm{20}}{\mathrm{2}×\mathrm{3}}}\approx\mathrm{10}.\mathrm{3}\:{m}/{s} \\ $$$${or} \\ $$$${t}=\frac{{d}}{{u}} \\ $$$${h}=\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}=\frac{{gd}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{2}} } \\ $$$$\Rightarrow{u}={d}\sqrt{\frac{{g}}{\mathrm{2}{h}}} \\ $$
Commented by ajfour last updated on 26/Dec/24
For both of you to watch this 13 min lecture of mine solving a rotation with slipping question, physics. Thank you.
https://youtu.be/Hg4sQD7xK9g?si=OMljkOWfCw2sIKV4