A) 8 sec
B) 10 sec
C) 12 sec
D) 16 sec
Correct Answer: C
Solution :
Given, \[y=\frac{-2}{3}{{t}^{2}}+16t+2\] Comparing it with\[y=\frac{1}{2}a{{t}^{2}}+ut\] we get \[u=16\] and \[\frac{1}{2}a=\frac{-2}{3}\], \[\Rightarrow \] \[a=\frac{-4}{3}\] Now \[v=u+at\] \[\therefore \] \[0=16-\frac{4}{3}t\] \[\Rightarrow \] \[t=\mathbf{12}\,\,\mathbf{sec}\]You need to login to perform this action.
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