A ball is thrown vertically from the top of a house 112 ft high. Its equation of motions is $s = - 16 t^{2} + 96 t$ where ' s ' is the directed distance of ball from the starting point at 't ' secs. Then the maximum height in feet attained by the ball and the time in seconds it takes to hit the ground are respectively

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128, 7

144, 7

144, 3

128, 3

answer is C.

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Detailed Solution

$\begin{array}{l} s = - 16 t^{2} + 96 t \\ \frac{d s}{d t} = - 32 t + 96 \\ 0 = - 32 t + 96 \\ \Rightarrow 32 t = 96 \\ \begin{array}{l} \Rightarrow t = \frac{96}{32} = \frac{24}{8} = 3 sec. \\ S = - 16 t^{2} + 96 t \\ S = - 16 (9) + 96 (3) \\ = - 144 + 288 \\ S = 144 \begin{array}{l} S = - 16 t^{2} + 96 t \\ 0 = - 16 t^{2} + 96 t \\ \Rightarrow 0 = 16 t (- t + 6) \\ \Rightarrow t = 0, t = 6 \\ t o t a l t i m e t o b e t a k e n t o r e a c h u p a n d d o w n = 6 \end{array} \end{array} \end{array}$