Answer:
\[\because \]velocity after\[5s=0+2\times 5=10m/s\]
height upto this point
\[=\frac{1}{2}a{{t}^{2}}=\frac{1}{2}\times 2\times {{(5)}^{2}}=25m\]
further height travelled by the body after
\[5s=\frac{{{0}^{2}}-{{10}^{2}}}{-2g}=5m\]
\[\therefore \] Total height\[=25+5=30m\]
Total time = time taken from A to B + time taken from B to max ht (C) and then to the ground.
\[t=5+{{t}^{1}}.\]How to get \[{{t}^{1}}\]? \[s=+25\text{ }m,\]
\[t=10m/s,\,\,t={{t}^{1}}=?,\,\,g=10m/{{s}^{2}}.\]
\[25=10{{t}^{1}}-\frac{1}{2}\times 10\times {{t}^{{{I}^{2}}}}\]
\[\Rightarrow 25=10{{t}^{1}}-5{{t}^{1}}\Rightarrow 5=2{{t}^{1}}-{{t}^{{{I}^{2}}}}\]
On solving this we get, \[{{t}^{1}}=1+\sqrt{6}s\]
\[\therefore \]total time \[t=5+1+\sqrt{6}=6+\sqrt{6}s\]
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