Question Number 180917 by anne1344 last updated on 19/Nov/22
Answered by mr W last updated on 19/Nov/22
$$\left.{c}\right) \\ $$$${h}_{{max}} =\frac{\left({v}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{\mathrm{2}{g}}=\frac{\left(\mathrm{15}×\mathrm{sin}\:\mathrm{20}°\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{9}.\mathrm{81}}=\mathrm{1}.\mathrm{34}{m} \\ $$$$\left.{d}\right) \\ $$$${t}=\frac{{v}\:\mathrm{sin}\:\theta}{{g}}=\frac{\mathrm{15}×\mathrm{sin}\:\mathrm{20}°}{\mathrm{9}.\mathrm{81}}=\mathrm{0}.\mathrm{52}\:{s} \\ $$