Question Number 213664 by 281981 last updated on 13/Nov/24
Answered by mr W last updated on 13/Nov/24
Commented by mr W last updated on 13/Nov/24
$$\frac{{x}}{\mathrm{20}−{y}}=\frac{\mathrm{5}}{{y}} \\ $$$${x}=\frac{\mathrm{100}}{{y}}−\mathrm{5} \\ $$$${u}=\frac{{dx}}{{dt}}=−\frac{\mathrm{100}}{{y}^{\mathrm{2}} }×\frac{{dy}}{{dt}} \\ $$$$\frac{{dy}}{{dt}}={v}=\sqrt{\mathrm{2}{gy}} \\ $$$$\Rightarrow{u}=−\frac{\mathrm{100}\sqrt{\mathrm{2}{gy}}}{{y}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{100}\sqrt{\mathrm{2}×\mathrm{10}×\mathrm{16}}}{\mathrm{16}^{\mathrm{2}} }\approx−\mathrm{7}\:{m}/{s} \\ $$
Answered by MathematicalUser2357 last updated on 14/Nov/24
$$\: \\ $$
Commented by mr W last updated on 14/Nov/24
$$\mathrm{1}\:{ft}\:=\:\mathrm{30}.\mathrm{48}\:{cm} \\ $$
Commented by Frix last updated on 14/Nov/24
আমিও দুঃখিত।