Question

A body is thrown vertically upwards with an initial velocity $u$ reaches a maximum height in $6s$. The ratio of the distance travelled by the body in the first second to the seventh second is

• A. 1:1
• B. 11:1
• C. 1:2
• D. 1:11

Answer 1

The correct option is B 11:1

height in the first second,
$h_1=u-\frac{g}{2}$
hight covered in the first second of downward journey
$h_2=\frac{g}{2}$
given $t_a=\frac{u}{g}=6;u=6g$

$v=0$ at max height
$v^2=u^2-2gs|t=\frac{u}{g}$
$\Rightarrow u^2=2g{H}_{max}|u=gt$

$u=\sqrt{2g{H}_{max}}|u=6g-\left(0\right)$
$s_1=u\left(1\right)-\frac{1}{2}g(1{)}^2=u-\frac{g}{2}.....(1)$
$s_7=\left[u\right(7)-\frac{1}{2}g(7{)}^2]-[u\left(6\right)-\frac{1}{2}g\left(6{)}^2\right]$
$=u-\frac{1}{2}g\left(13\right).....\left(2\right)$
$\frac{s_1}{s_7}=\frac{u-\frac{g}{2}}{u-\frac{g}{2}\left(13\right)}=\frac{6g-\frac{g}{2}}{6g-\frac{13g}{2}}=\frac{11}{-1}.....\left(3\right)$
So, $\frac{s_1}{s_7}=\frac{11}{1}$