Question

A body is projected vertically upwards with a velocity $u$. It crosses a point in its journey at a height $h$ meter twice, just after $1$ and $7$ seconds .The value of $u$ in $m{s}^{-1}$ is $(g=10m{s}^{-2})$

• A. $50$
• B. $40$
• C. $30$
• D. $20$

Answer 1

Answer: (B)

$40$

$h=ut-\frac{1}{2}g{t}^2$
$h=u\left(1\right)-\frac{1}{2}g=u-\frac{g}{2}.....\left(1\right)$
$h=u\left(7\right)-\frac{1}{2}g(7{)}^2=7u-\frac{49g}{2}.....(2)$
Equating (1) & (2)
$u-\frac{g}{2}=7u-\frac{49g}{2}$
$24g=6u$
$u=4g$
$\Rightarrow u=40m/s$