Question

A body projected vertically upward with a velocity $v$ at $t=0$ is found at a height $h$ after $1s$ and a further $6s$, it is found at the same height $(g=10m{s}^{-2})$. Then

• A. $h$ is $30m$
• B. $v$ is $40m{s}^{-1}$
• C. maximum height is $80m$
• D. distance moved in $5th$ second is $5m$

Answer 1

Answer: (B), (C), (D)

maximum height is $80m$
$v$ is $40m{s}^{-1}$
distance moved in $5th$ second is $5m$

Let that height be $h$.
$Thus,h=v\left(1\right)-\frac{1}{2}\times 10\times {\left(1\right)}^2=v\left(7\right)-\frac{1}{2}\times 10\times {\left(7\right)}^{2}Thisgivesv=40m/sandh=35mNow,maxheightmeansvelocitybecomes0Hence,H_{max}=\frac{0^2-{40}^2}{2\times (-10)}=80m0=40-10tgivestimetoreachthisheightas4seconds.Thusin5thseconditstartsfromzerovelocity.Thusdistance,s=0+\frac{1}{2}\times 10\times 1^2=5m$