A) 2s
B) 4 s
C) 3 s
D) 3.5 s
E) 2.5 s
Correct Answer: E
Solution :
\[s=80t-16{{t}^{2}}\] On differentiating w. r. t. t, we get \[\frac{ds}{dt}=80-32t\] Again differentiating, we get, \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=-32\] Put\[\frac{ds}{dt}=0\]for maxima or minima \[\Rightarrow \] \[t=2.5\] \[{{\left( \frac{{{d}^{2}}s}{d{{t}^{2}}} \right)}_{t=2.5}}=-32<0\] \[\therefore \]The particle required 2.5 s to reach the maximum height.
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