question_answer

A particle is projected vertically up with \[20\,m{{s}^{-1}}\] then travelled by it in initial 3 sec. will be : \[(g=10m{{s}^{-2}})\]

A)  15m                           

B)  25m

C)  20m                           

D)  30m

Correct Answer: B

Solution :

\[t=\frac{v}{g}=\frac{20}{10}=2\sec \] \[h=\frac{1}{2}g{{t}^{2}}=\frac{1}{2}\times 10\times {{1}^{2}}=5\] \[h=\frac{{{v}^{2}}}{2g}=20\] distance = path length = 20 + 5 = 25m