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2.Motion in Straight Line
hard
From the top of a $64\, ft$ high tower, a stone is thrown upwards vertically with the velocity of $48\, ft/s$. The greatest height (in $ft$) attained by the stone, assuming the value of the gravitational acceleration $g\, = 32\, ft/s^2$, is
A
$128$
B
$88$
C
$112$
D
$100$
Solution
$\begin{array}{l}
Let'u'\,be\,the\,velocity\,\\
\therefore u = 48m/s,\,Given,\,g = 32\\
At\,{\rm{maximum}}\,height\,v = 0\\
Now,\,we\,know\,{v^2} = {u^2} – 2gh\\
\Rightarrow 0 = {\left( {48} \right)^2} – 2\left( {32} \right)h \Rightarrow h = 36\\
Maximum\,height\, = 36 + 64 = 100\,mt
\end{array}$
Standard 11
Physics
