A) \[\frac{h}{9}\] m from the ground
B) \[\frac{7h}{9}\] m from the ground
C) \[\frac{8h}{9}\]m from the ground
D) \[\frac{17h}{18}\] m from the ground
Correct Answer: C
Solution :
| [c] As, initial velocity of ball is zero. |
| \[\because \ \ \ \ \ \ \ \ \ \ \ \ \ s=ut+\frac{1}{2}g{{t}^{2}}\] |
| \[\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ h=\frac{1}{2}g{{T}^{2}}\] |
| \[\operatorname{In}=\frac{T}{3}\sec ,\] |
| \[{{h}_{1}}=\frac{1}{2}g{{\left( \frac{T}{3} \right)}^{2}}=\frac{1}{2}g\left( \frac{{{T}^{2}}}{9} \right)\] |
| From Eqs. (i) and (ii), we get |
| \[{{h}_{1}}=\frac{h}{9}\] |
| Distance of ball from ground =\[h=\frac{h}{9}=\frac{8h}{9}\operatorname{m}\] |
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