Question Number 68272 by peter frank last updated on 08/Sep/19
Answered by Kunal12588 last updated on 08/Sep/19
![H_(max) =maximum height R=horizontal range H_(max) =h=(u_y ^2 /(2g))⇒u_y =(√(2gh)) R=r=((2u_x u_y )/g)⇒u_x =((gr)/(2u_y ))=((gr)/(2(√(2gh))))=(r/2)(√(g/(2h))) u=(√(u_x ^2 +u_y ^2 )) =(√(((r^2 g)/(8h))+2gh)) =(√(2g((r^2 /(16h))+h))) =[2g(h+(r^2 /(16h)))]^(1/2)](https://www.tinkutara.com/question/Q68275.png)
$${H}_{{max}} ={maximum}\:{height} \\ $$$${R}={horizontal}\:{range} \\ $$$${H}_{{max}} ={h}=\frac{{u}_{{y}} ^{\mathrm{2}} }{\mathrm{2}{g}}\Rightarrow{u}_{{y}} =\sqrt{\mathrm{2}{gh}} \\ $$$${R}={r}=\frac{\mathrm{2}{u}_{{x}} {u}_{{y}} }{{g}}\Rightarrow{u}_{{x}} =\frac{{gr}}{\mathrm{2}{u}_{{y}} }=\frac{{gr}}{\mathrm{2}\sqrt{\mathrm{2}{gh}}}=\frac{{r}}{\mathrm{2}}\sqrt{\frac{{g}}{\mathrm{2}{h}}} \\ $$$${u}=\sqrt{{u}_{{x}} ^{\mathrm{2}} +{u}_{{y}} ^{\mathrm{2}} } \\ $$$$=\sqrt{\frac{{r}^{\mathrm{2}} {g}}{\mathrm{8}{h}}+\mathrm{2}{gh}} \\ $$$$=\sqrt{\mathrm{2}{g}\left(\frac{{r}^{\mathrm{2}} }{\mathrm{16}{h}}+{h}\right)} \\ $$$$=\left[\mathrm{2}{g}\left({h}+\frac{{r}^{\mathrm{2}} }{\mathrm{16}{h}}\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$
Commented by peter frank last updated on 08/Sep/19
$${thank}\:{you} \\ $$