Question

Aball is droped from the top of the tower of height h .It covers a distance of h/2 in the last second of its motion .How long does the ball remain in air .(Take g = 10 ms${}^{-2}$)

• A. $\sqrt{2}$ s
• B. $(2+\sqrt{2}$) s
• C. $2s$
• D. None of the above

Answer 1

Answer: (B)

$(2+\sqrt{2}$) s

Let time during which ball remain in the air $t$,

From the kinematics,

$s=ut+\frac{1}{2}g{t}^2$ ($u=0m/s$)

$h=\frac{g{t}^2}{2}$. . . . .(1)

The distance travelled by the ball in $n^{th}sec$ is given by

$S_n=u+\frac{g}{2}(2n-1)$

at $n=(t-1)sec$

$\frac{h}{2}=\frac{g}{2}\left[2\right(t-1)-1]$

$h=g(2t-3)$. . . . . (2)

By solving equation (1) and (2), we get

$\frac{g{t}^2}{2}=g(2t-3)$

$t^2-4t+6=0$ (Quadratic equation)

$t=\frac{\sqrt{2}}{\sqrt{2}-1}$

By rationalisig,

$t=\frac{\sqrt{2}}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}$

$t=(2+\sqrt{2})s$

The correct option is B.