# A fire cracker is fired and it rises to a height of 1000 m. Find the velocity by which it was released and the time taken by it to reach the highest point (take a = 10 m/s?).

(1) h=1000m =S

v= 0 m/s (at highest point)

g= 10 m/s

^{2}( taken to be negative according to sign convention)

Using the 3

^{rd}equation of motion,

${v}^{2}-{u}^{2}=2aS\phantom{\rule{0ex}{0ex}}\Rightarrow 0-{u}^{2}=-2\times 10\times 1000\phantom{\rule{0ex}{0ex}}\Rightarrow {u}^{2}=20000\phantom{\rule{0ex}{0ex}}\Rightarrow u=141.42m/s$

(2) u=141.42 m/s (as calculated above)

1

^{st}equation of motion,

$V=u+at\phantom{\rule{0ex}{0ex}}\Rightarrow 0=141.42-10t\phantom{\rule{0ex}{0ex}}\Rightarrow 10t=141.42\phantom{\rule{0ex}{0ex}}\Rightarrow t=14.142sec\phantom{\rule{0ex}{0ex}}\Rightarrow t=14.1sec\left(approx\right)$

velocity by which it is released is 141.42m/s

and time taken to reach the highest point =14.1 sec

Regards,

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