Question

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

• A. $h$ is $30m$
• B. $v$ is $40m/s$
• C. maximum height is $80m$
• D. distance moved in $5th$ second is $5m$

Answer 1

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

$v$ is $40m/s$
maximum height is $80m$
distance moved in $5th$ second is $5m$

$\theta =90\Rightarrow t=\frac{2v}{g}\&H=\frac{v^2}{2g}$
Total time will be 8 seconds as it takes 1s to reach $h$ from initial position and then 6s to return back to $h$ and 1s again to reach the initial position.
$t=8=\frac{2v}{g}\Rightarrow v=40m/s$
$H=\frac{v^2}{2g}\Rightarrow H=\frac{40\times 40}{2\times 10}=80m$
Now, Velocity will be 0 at top also it will reach there at $t=4s$ as total time is 8s
$s=ut+\frac{1}{2}a{t}^2$
$s=\frac{1}{2}\times 10\times 1=5m$