A) \[t={{t}_{1}}-{{t}_{2}}\]
B) \[t=\frac{{{t}_{1}}+{{t}_{2}}}{2}\]
C) \[t=\sqrt{{{t}_{1}}{{t}_{2}}}\]
D) \[t=\sqrt{t_{1}^{2}-t_{2}^{2}}\]
Correct Answer: C
Solution :
From equation of motion, we have \[s=ut+\frac{1}{2}g{{t}^{2}}\] where, \[u\]is initial velocity, g the acceleration due to gravity and \[t\] the time. For upward motion \[h=-u{{t}_{1}}-\frac{1}{2}gt_{1}^{2}\] ?(i) For downward motion \[h=-u{{t}_{2}}+\frac{1}{2}gt_{2}^{2}\] ?(ii) Multiplying Eq. (i) by \[{{t}_{2}}\]and Eq. (ii) by \[{{t}_{1}}\]and subtracting Eq. (ii) by Eq. (i), we get \[h({{t}_{2}}-{{t}_{1}})=\frac{1}{2}g{{t}_{1}}{{t}_{2}}({{t}_{2}}-{{t}_{1}})\] \[h=\frac{1}{2}g{{t}_{1}}{{t}_{2}}\] ?(iii) When stone is dropped from rest\[u=0,\] reaches the ground in t second. \[\therefore \] \[h=\frac{1}{2}g{{t}^{2}}\] ?(iv) Equating Eqs. (iii) and (iv), we get \[\frac{1}{2}g{{t}^{2}}=\frac{1}{2}g{{t}_{1}}{{t}_{2}}\] \[\Rightarrow \] \[{{t}^{2}}={{t}_{1}}{{t}_{2}}\Rightarrow \sqrt{{{t}_{1}}t & {{ & }_{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
