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) \[\frac{17h}{9}m\]from the ground
Correct Answer: C
Solution :
Second law of motion \[S=ut+\frac{1}{2}g{{t}^{2}}\] \[h=0+\frac{1}{2}g{{t}^{2}}\] \[T=\sqrt{\left( \frac{2h}{g} \right)}\] \[t=\frac{T}{3}s\] \[S=0=\frac{1}{2}g{{\left( \frac{T}{3} \right)}^{2}}\] \[S=\frac{1}{2}g.\frac{{{T}^{2}}}{9}\] \[S=\frac{g}{18}\times \frac{2h}{g}\] \[S=\frac{h}{9}m\] Hence, the position of ball from the grou \[=h-\frac{h}{9}=\frac{8h}{9}m\]
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