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A particle is projected vertically upwards with an initial velocity of `40 m//s.` Find the displacement and distance covered by the particle in `6 s.` Take `g= 10 m//s^2.` |
Answer» Correct Answer - A In 4s, it reaches upto the highest point and then changes its direction of motion. `s=ut+1/2 at^2=(40)(6)+1/2(-10)(6)^2` `=60 m` `d=|s|_(0-4)+|s|_(4-6)=|u^2/(2g)|+|1/2 g(t-t_0)^2|` `=(40)^2/(2xx10)+1/2xx10xx(6-4)^2=100 m` |
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