Question

A small platform starts moving from rest from ground in upward direction with uniform acceleration $5m{s}^{-2}$. At time $t=2s$, a stone is projected horizontally in frame of platform with speed $10m{s}^{-1}$. The maximum height attained by the stone with respect to the ground is (take $g=10m/s^2$)

• A. $10m$
• B. $15m$
• C. $25m$
• D. $20m$

Answer 1

The correct option is B $15m$

After $t=2s$ the velocity of the platform in the upward direction is: $v=u+at$

$\Rightarrow v=0+5\left(2\right)=10$ m/s

Height of platform from ground at this time is:

$h=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}\left(5\right)\left(4\right)=10$m

From ground frame the velocity of stone is vertical direction:

$v_y=10$ m/s

and velocity of stone in horizontal direction:

$v_x=10$ m/s

So motion of stone is a projectile with angle ${\tan }^{-1}\frac{10}{10}={45}^o$ and initial velocity $=10\sqrt{2}$ m/s

So max height from $10$m level with respect to ground is:

$h_{max}=\frac{u^2{\sin }^2\theta }{2g}=\frac{\left(10\sqrt{2}{)}^2\right(1/2)}{2\left(10\right)}=\frac{10}{2}=5m$

Hence net max height from ground $=10+5=15$m.