A) \[80\,\,m\]
B) \[100\,\,m\]
C) \[60\,\,m\]
D) \[40\,\,m\]
Correct Answer: D
Solution :
For the first stone,\[{{u}_{1}}=0\] Let the first stone pick up a speed of \[30\,\,m\,\,{{s}^{-1}}\]after\[{{t}_{1}}\sec \]. Then,\[{{v}_{1}}={{u}_{1}}+g{{t}_{1}}\] \[\Rightarrow \]\[{{t}_{1}}=30/10=3\,\,s\]. Total distance travelled by the first stone in\[3\] seconds is:\[{{s}_{1}}={{u}_{1}}{{t}_{1}}+\frac{1}{2}g{{t}_{1}}^{2}\] \[=0+\frac{1}{2}\times 10\times {{(3)}^{2}}=45\,\,m\] The second stone is dropped after \[2\] seconds. This means that it only travelled for \[1\,\,\sec \] by the time the first stone reaches a speed of \[30\] m/second. Distance travelled by the second stone in \[1\] second. \[\Rightarrow \]\[{{s}_{2}}={{u}_{2}}{{t}_{2}}+\frac{1}{2}gt_{2}^{2}=0+\frac{1}{2}\times 10\times {{(1)}^{2}}\] \[=5\,\,m\]. \[\therefore \]Distance between both the stones \[={{s}_{1}}-{{s}_{2}}=45-5=40\,\,m\]
You need to login to perform this action.
You will be redirected in
3 sec
