Question

A particle of mass $2m$ is projected at an angle of ${45}^o$ with horizontal with a velocity of $20\sqrt{2}m/s$. After $1s$ explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. Find the maximum height attained by the other part. Take $g=10m/s^2$:

• A. $25m$
• B. $50m$
• C. $15m$
• D. $35m$

Answer 1

The correct option is D $35m$

Height moved by particle in $1sec$,

$h=ut-\frac{1}{2}g{t}^2$

$=20\times 1-\frac{1}{2}\times 10\times 1$

$=15m$

After $1sec$ vertical speed is $v$,

$v-u=-gt$

$v-20=-10\times 1$

$v=10m/s$

angle $\theta$ at height $h$ with horizontal is,

$\tan \theta =\frac{10}{20}=\frac{1}{2},\sin \theta =\frac{1}{\sqrt{5}}$

before explosion velocity$=\sqrt{{10}^2+{20}^2}$

$=\sqrt{500}=10\sqrt{5}$

After explosion $m$ is get $\frac{1}{2}m$, So, by conservation of energy velocity will be double$=20\sqrt{5}$

Height travelled after explosion$=\frac{u^2{\sin }^2\theta }{2g}$

$=\frac{(20\sqrt{5}{)}^2\times {\sin }^2\theta }{2g}$

$=\frac{400\times 5\times \frac{1}{5}}{2\times 10}$

$=20$

Total height$=20+15=35cm$