Question

To what time you remain air borne(free fall) until you return to saddle

• A. 0.582 s
• B. 0.528 s
• C. 0.483 s
• D. none

Answer 1

Answer: (B)

0.528 s

The motion of the man is upward with velocity $v=0.25(2\pi \times 1.5)$
$v=0.75\pi m{s}^{-1}$ and falls with accelerating g.
Time taken to return to leaving point $=\frac{2v}{g}=\frac{2\times \left(0.75\pi \right)}{g}=\frac{1.5}{\pi }=0.48s$.
In that time, saddle has not reached to him as $T=\frac{1}{1.5}=0.666s$.
Therefore,man will settle down on saddle again when the two are at same position. When man is at extreme position during return(0.25m), saddle is at $y=0.25\cos 3\pi \left(0.48\right)=-0.047m$ from mean position. Therefore, the saddle reaches to him when $a=g$
$g=y(2\pi f{)}^2$
or $y=\frac{1}{9}m;y=y_0\sin \omega t$
$t=0.048;$ total time $=0.48+.048$
$=0.528s$