Question

A platform is moving upwards with a constant acceleration of $2m{s}^{-2}$. At time $t=0$, a boy standing on the platform throws a ball upwards with a relative speed of $8m{s}^{-1}$. At this instant, platform was at the height of 4 m from the ground and was moving with a speed of $2m{s}^{-1}$.
Take g = 10 m s${}^{-2}$. Find
a. when and where the ball strikes the platform.
b. the maximum height attained by the ball from the ground.
c. the maximum distance of the ball from the platform.

Answer 1

We solve the problem in reference frame of platform,

${\overline{v}}_{ball/platform}=8\hat{j}$

${\overline{a}}_{P/E}=2\hat{j}$ and ${\overline{a}}_{B/E}=-g\hat{j}$

${\overline{a}}_{rel}={\overline{a}}_{B/P}=-12\hat{j}$

By $s_{rel}=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2$

$0=8\times t-\frac{1}{2}\times 12t^2$

$\Rightarrow t=\frac{4}{3}s$

Total Time $2+\frac{4}{3}=\frac{10}{3}s$

Displacement of platform in 10/3s

$=4+2\times \frac{4}{3}+\frac{1}{2}\times 2\times {\left(\frac{4}{3}\right)}^2=\frac{76}{9}m$

$b.$

${\overline{v}}_{B/E}=10\hat{j}$ and ${\overline{a}}_{B/E}=-10\hat{j}$

by $v^2=u^2+2as$ w,r,t earth,

$(0{)}^2=(10{)}^2-2\left(10\right)s_1\Rightarrow s_1=5m$

$H_{max}=5+4=9m$

$c.$

Also platform frame, $v^2=u^2+2as$

$(0{)}^2=(8{)}^2+2(-12)s\Rightarrow s=\frac{8}{3}m$