A) \[h/9\] metres from the ground
B) \[7h/9\,\,m\] m from the ground
C) \[8h/9\] metres from the ground
D) \[17h/18\,\,m\] from the ground
Correct Answer: C
Solution :
\[{{h}^{1}}=\frac{1}{2}\times g\times \frac{{{T}^{2}}}{9}\]and\[T=\sqrt{\frac{2h}{g}}\] \[\therefore \]\[{{h}^{1}}=\frac{1}{{}}\times \times \frac{h}{\times 9},\,\,{{h}^{1}}=\frac{h}{9}\] From ground, the position of ball is \[\left( h-\frac{h}{9} \right)m=\frac{8h}{9}m\]
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