Question Number 210753 by peter frank last updated on 18/Aug/24
Answered by mr W last updated on 18/Aug/24
Commented by mr W last updated on 18/Aug/24
$$\left.{a}\right) \\ $$$${v}=\mathrm{30}\:{m}/{s} \\ $$$${s}={vt}\:\Rightarrow{t}=\frac{{s}}{{v}} \\ $$$$\mathrm{100}+{s}={u}\:\mathrm{cos}\:\theta\:{t}\: \\ $$$$\Rightarrow{u}\:\mathrm{cos}\:\theta=\frac{\left(\mathrm{100}+{s}\right){v}}{{s}}\:\:\:\:…\left({i}\right) \\ $$$$\mathrm{0}={u}\:\mathrm{sin}\:\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}\: \\ $$$$\Rightarrow{u}\:\mathrm{sin}\:\theta=\frac{{gt}}{\mathrm{2}}=\frac{{gs}}{\mathrm{2}{v}}\:\:\:…\left({ii}\right) \\ $$$$\left({ii}\right)/\left({i}\right): \\ $$$$\mathrm{tan}\:\theta=\frac{{gs}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{100}+{s}\right){v}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{9}.\mathrm{81}{s}^{\mathrm{2}} }{\mathrm{1800}\left(\mathrm{100}+{s}\right)} \\ $$$$\Rightarrow\mathrm{9}.\mathrm{81}\sqrt{\mathrm{3}}{s}^{\mathrm{2}} −\mathrm{1800}{s}−\mathrm{180000}=\mathrm{0} \\ $$$$\Rightarrow{s}=\frac{\mathrm{900}+\sqrt{\mathrm{900}^{\mathrm{2}} +\mathrm{180000}×\mathrm{9}.\mathrm{81}\sqrt{\mathrm{3}}}}{\:\mathrm{9}.\mathrm{81}\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\approx\mathrm{168}.\mathrm{7}\:{m} \\ $$$${AC}=\mathrm{100}+{s}=\mathrm{268}.\mathrm{7}\:{m}\:\checkmark \\ $$$$\left.{b}\right) \\ $$$${at}\:{point}\:{C}\:{the}\:{shell}\:{has}\:{the}\:{same} \\ $$$${velocity}\:{as}\:{at}\:{point}\:{A}. \\ $$$${u}_{{C}} ={u}=\frac{{gs}}{\mathrm{2}{v}\:\mathrm{sin}\:\theta}=\frac{\mathrm{9}.\mathrm{81}×\mathrm{168}.\mathrm{7}}{\mathrm{2}×\mathrm{30}×\mathrm{0}.\mathrm{5}} \\ $$$$\:\:\:\:\:\:=\mathrm{55}.\mathrm{2}\:{m}/{s}\:\checkmark \\ $$
Commented by peter frank last updated on 19/Aug/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{mr}\:\mathrm{W} \\ $$