Question

A body is thrown upwards with velocity 40 m s${}^{-1}$ and it covers 5 m in the last second of its upward journey. If the same body is thrown upwards with velocity 80 m s${}^{-1}$, what distance will it travel in the last second of upward journey? (Take g = 10 m s${}^{-2}$)

• A. $5m$
• B. $10m$
• C. $15m$
• D. $20m$

Answer 1

The correct option is A $5m$

Now,

$v^2=u^2-2gs0=(80{)}^2-(2\left)\right(10\left)\right(s)0=6400-20s20s=6400s=320metrev=u+gt0=80-10\pi t0=80-10t10t=80t=8seconds$

In last second

$S_{20}=u+\left(\frac{1}{2}\right)g(2t-1)=80-\left(\frac{1}{2}\right)\left(10\right)(2\times 8-1)=80-5\left(15\right)=80-75=5metres$

A is the correct option.