| A ball is dropped from a high rise platform at \[t=0\] starting from rest. After 6s another ball is thrown downwards from the same platform with a speed v. The two balls meet at \[t=18s\]. [AIPMT (S) 2010] |
| What is the value of v? (take \[g=10\text{ }m{{s}^{-2}}\]) |
A) \[74\,\,m{{s}^{-2}}\]
B) \[55\,\,m{{s}^{-1}}\]
C) \[40\,m{{s}^{-1}}\]
D) \[60\,m{{s}^{-1}}\]
Correct Answer: A
Solution :
| For first ball, \[u=0\] |
| \[\therefore \] \[{{s}_{1}}=\frac{1}{2}gt_{1}^{2}=\frac{1}{2}\times g{{(18)}^{2}}\] |
| For second ball, initial velocity \[=v\] |
| \[\therefore \] \[{{s}_{2}}=v{{t}_{2}}+\frac{1}{2}g{{t}^{2}}\] |
| \[{{t}_{2}}=18-6=12s\] |
| \[\Rightarrow \] \[{{s}_{2}}=v\times 12+\frac{1}{2}g{{(12)}^{2}}\] |
| Here, \[{{s}_{1}}={{s}_{2}}\] |
| \[\frac{1}{2}g{{(18)}^{2}}=12v+\frac{1}{2}g{{(12)}^{2}}\] |
| \[\Rightarrow \] \[v=74\,\,m{{s}^{-1}}\] |
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