Question

A projectile is fired from a gun at an angle of ${45}^o$ with the horizontal and with a speed of $20$ m/s relative to ground. At the highest point in its flight the projectile explodes into two fragments of equal mass. One fragment comes at rest just after explosion. How far from the gun does the other fragment land, assuming a horizontal ground? Take $g=10m/s^2$?

Answer 1

As the first part comes to the rest means, then it will fall under gravity directly.

First of all maximum height,

$H=\frac{u²sin²\theta }{g}$

$H=\frac{400\times \left(\frac{1}{2}\right)}{2\times 10}$

$H=10m.$

As initially the full projectile was supposed to land at A but due to explosion first part land at A', so to make the centre of mass lies at A the second part should lie at B.

f they don't break (explode) range OA

$OA=\frac{u²sin2\theta }{g}$

$OA=400\times sin45\times \frac{2}{10}$

$OA=400\times \frac{sin90°}{10}$

$OA=40m.$

Therefore ,

$A{A}^{\prime }=\frac{OA}{2}$

$A{A}^{\prime }=20m.$

As the first part is at 20 m from A in left side being equal mass the second part will be at 20 m right to A.

From gun it is $OB=40+20m=60m.$

so gun does the other fragment land at $60m.$