Answer:
If h is the height of the tower, displacement\[=-h\] So using \[\therefore \,s=ut+\frac{1}{2}a{{t}^{2}}\] \[-h=19.6\,\,t-\frac{1}{2}\times 9.8\,{{t}^{2}}\] Since\[u=19.6\,m{{s}^{-1}}\]and it moves up initially. i.e. \[-58.8=19.6\,\,t-4.9\,\,{{t}^{2}}\] \[4.9\,{{t}^{2}}-19.6\,\,t-4.9\,\,{{t}^{2}}-58.8={{t}^{2}}-4t-12=0\] \[\Rightarrow (t-6)(t+2)=0\,\,\,\therefore \,\,t=6s\]
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