A) h/9 meters from the ground
B) 7h/9 meters from the ground
C) 8h/9 meters from the ground
D) 17h/18 meters from the ground
Correct Answer: C
Solution :
\[\because \ \ h=ut+\frac{1}{2}g{{t}^{2}}\]\[\Rightarrow h=\frac{1}{2}g{{T}^{2}}\] After \[\frac{T}{3}\] seconds, the position of ball, \[{h}'=0+\frac{1}{2}g{{\left( \frac{T}{3} \right)}^{2}}=\frac{1}{2}\times \frac{g}{9}\times {{T}^{2}}\] \[h'=\frac{1}{2}\times \frac{g}{9}\times {{T}^{2}}\]\[=\frac{h}{9}m\] from top \ Position of ball from ground \[=h-\frac{h}{9}=\frac{8\ h}{9}m.\]You need to login to perform this action.
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