Question

A massive vertical wall is approaching a man at a speed u. When it is at a distance of $10$m, the man throws a ball with speed $10m/s$ at an angle of ${37}^o$ which after completely elastic rebound reaches back directly into his hands. Find the velocity u of the wall.

Answer 1

Total time in vertical motion $=$ Total time in horizontal motion
$u_x=8m^{-1},u_y=6m{s}^{-1}$

time taken to collide against wall $=\frac{10}{8+u}=t_1$

In that $t_1$ time wall will more by $t_1\times u=\frac{u\times 10}{8+u}$

$\therefore$ Distance covered in return journey$=10-\frac{u\times 10}{8+u}$

Again, velocity while restoring $\Rightarrow (e=1)$
can be calculated

$\Rightarrow$velocity of approach $=$ velocity of representation

let, return velocity be $v$

$v-u=u+8$

$v=2u+8$

Time taken while restoring $=\frac{80}{(8+u)(2u+8)}=t_2$

$\therefore t_1+t_2=$ Time taken in verticle journy

If time taken to go up is $t$

then, $\left(10\sin {37}^o\right)-gt=0$

$\Rightarrow 10\times \frac{3}{5}=10t=0$

$\Rightarrow t=33/5$

$\therefore$ time of go return $=2\times 3/5$

$=6/5$

$=\frac{10}{u+8}+\frac{80}{(u+8)(2u+8)}=\frac{6}{5}$

$u=13/3$ Ans.