A) \[\frac{1}{2}\]
B) 2
C) \[\frac{1}{4}\]
D) 4
Correct Answer: A
Solution :
| For \[A\,to\,B\] |
| \[u=0;s=-(H-h);\,a=-g;\,t=?\] |
| \[s=ut+\frac{1}{2}a{{t}^{2}}\Rightarrow -(H-h)=\frac{1}{2}(-g){{t}^{2}}\]\[\Rightarrow t={{\left[ \frac{2\left( H-h \right)}{g} \right]}^{1/2}}\] |
| For B to C vertical motion |
| \[{{u}_{y}}=0;\,{{s}_{y}}=-h;{{a}_{y}}=-g\] |
| \[s=ut+\frac{1}{2}a{{t}^{2}}\]\[-h=\frac{1}{2}(-g){{t}^{{{'}^{2}}}}\Rightarrow {{t}^{'}}=\sqrt{\frac{2h}{g}}\] |
| Total time of fall \[T=t+{{t}^{'}}\] |
| \[T={{\left[ \frac{2\left( H-h \right)}{g} \right]}^{1/2}}+{{\left[ \frac{2h}{g} \right]}^{1/2}}\] |
| Note: for finding the maximum time, using the concept of differentiation |
| We have \[\frac{dT}{dh}=0\] |
| \[\Rightarrow \frac{d}{dt}{{\left[ \frac{2\left( H-h \right)}{g} \right]}^{1/2}}+\frac{d}{dt}{{\left[ \frac{2h}{g} \right]}^{1/2}}=0\]\[\Rightarrow \frac{1}{2}{{\left[ \frac{2\left( H-h \right)}{g} \right]}^{-1/2}}\times \left( \frac{-2}{g} \right)+\frac{1}{2}{{\left[ \frac{2h}{g} \right]}^{-1/2}}\left( \frac{2}{g} \right)=0\]\[\Rightarrow {{\left[ \frac{2\left( H-h \right)}{g} \right]}^{-1/2}}={{\left( \frac{2h}{g} \right)}^{-1/2}}\]\[\Rightarrow \frac{2\left( H-h \right)}{g}=\frac{2h}{g}\Rightarrow H-h=h\Rightarrow \frac{h}{H}=\frac{1}{2}\] |
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