Question

A toy plane $P$ starts flying from point $A$ along a straight horizontal line $20m$ above ground level starting with zero initial velocity and acceleration $2m/s^2$ as shown. At the same instant, a man $P$ throws a ball vertically upwards with initial velocity ${}^{\prime }{u}^{\prime }$. Ball touches (coming to rest) the base of the plane at point $B$ of plane's journey when it is vertically above the mans. ${}^{\prime }{s}^{\prime }$ is the distance of point $B$ from point $A$. Just after the contact of ball with the plane, acceleration of plane increases to $3m/s^2$. Find:
(i) Initial velocity ${}^{\prime }{u}^{\prime }$ of ball.
(ii) Distance ${}^{\prime }{s}^{\prime }$
(iii) Distance between man and plane when the man catches the ball black. ($g=10m/s^2)$ (Neglect the height of man).

Answer 1

Time taken by ball to reach $B$ from $O=t=$ times taken by plane to each $B$ from $A$

(i) speed of ball at $B=O=V$

$v^2=u^2-2gs\Rightarrow U^2=2\times 10\times 20\Rightarrow u=20m/s$

(ii) $v=u-gt\Rightarrow u=20-\left(10t\right)\Rightarrow t=2sec$

for plane $S=ut+\frac{1}{2}a{t}^2=0+\frac{}{}\times 2\times 2^2=4m$

$\Rightarrow s=4m$

(iii) time taken by ball to return back $=2sec$

Let $BC=d$

speed of plane at $B=u+at=0+(2\times 2)=4m/s$

$d=(4\times 2)\to \frac{1}{2}\times 3\times 2^2=8+6=14m$

distance between man and plane $=\sqrt{{20}^2+{14}^2}=24.4m$