Question

A body is allowed to fall from a height of $98$ m before hitting the ground the distance travelled by it in the last second of motion $(g=9.8m{s}^{-2})$ is

• A. $38m$
• B. $40m$
• C. $50m$
• D. $29m$

Answer 1

The correct option is A $38m$

$h=98$m

fall $\Rightarrow 4=0$

Using $h=ut+\frac{1}{2}g{t}^2$

$\Rightarrow h=\frac{1}{2}g{t}^2$

$\Rightarrow t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 98}{9.8}}=\sqrt{20}\sec =4.47\sec$

or $t\approx 4.5\sec$

$\Rightarrow t-1=4.5-1=3.5\sec$

Distance fall in $3.5\sec$ will be

$h=\frac{1}{2}g{t}^2=\frac{1}{2}\times 9.8\times (3.5{)}^2=60$ meter

So required distance $=98-60=38$ meter.