Question

A particle of mass $2m$ is projected at an angle ${45}^{\circ }$ with the horizontal with a velocity of $20\sqrt{2}m/s$. After $1s$, explosion takes place and the particle is broken into the two equal pieces. As a result of explosion, one part comes to rest. The maximum height from the ground attained by the other part is

• A. $50m$
• B. $25m$
• C. $40m$
• D. $35m$

Answer 1

Answer: (D)

$35m$

Given: Initial velocity $u_0=20\sqrt{2}m/s$; angle of projection $\theta ={45}^{\circ }$
Therefore horizontal and vertical components of initial velocity are
$u_x=20\sqrt{2}\cos {45}^{\circ }=20m/s$
and $u_y=20\sqrt{2}\sin {45}^{\circ }=20m/s$
After $1s$, horizontal component remains unchanged while the vertical component becomes
$v_y=u_y-gt$
Due to explosion, one part comes to rest.
Hence, from the conservation of linear momentum, vertical component of second part will become $v_y^{\prime }=20m/s$.
Therefore, maximum height attained by the second part will be
$H=h_1+h_2$
Using, $h=ut+\frac{1}{2}a{t}^2$
$\Rightarrow h_1=(20\times 1)-\frac{1}{2}\times 10\times (1{)}^2=15m$
$a=g=10m/s^2$
$h_2=\frac{v_y^{\prime 2}}{2g}=\frac{(20{)}^2}{2\times 10}=20m$
$H=20+15=35m$.