A) 980 m
B) 735 m
C) 490 m
D) 245 m
Correct Answer: B
Solution :
Velocity when the engine is switched off \[v=19.6\times 5=98\text{ m}{{\text{s}}^{-1}}\text{ }{{\text{h}}_{\max }}={{h}_{1}}+{{h}_{{}}}\] Where \[{{\text{h}}_{\text{1}}}\text{=}\frac{\text{1}}{\text{2}}\text{a}{{\text{t}}^{\text{2}}}\text{, }{{\text{h}}_{\text{2}}}\text{=}\frac{{{\text{v}}^{\text{2}}}}{\text{2a}}\] \[{{h}_{\max }}=\frac{1}{2}\times 19.6\times 5\times 5+\frac{98\times 98}{2\times 9.8}=73.5\text{ m}\]
You need to login to perform this action.
You will be redirected in
3 sec
