A projectle is fired from a gun at an angle $45^{o}$ with a speed of m/s relative to ground. At the highest point in its flight the 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 $g = 10 m / s^{2} ?$

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Solution

First of all maximum height
$H = \frac{u^{2} S i n^{2} θ}{g}$
$H = \frac{(400 \times \frac{1}{2})}{2} \times 10$

H = 10m

if they don't explode range OA

$O A = \frac{u^{2} s i n^{2} θ}{g} = O A = 400 \times \frac{s i n 90^{\circ}}{10} = 40 m$

Therefore $A A = \frac{O A}{2} = 20 m$

From gun it is $O B = 40 + 20 m$
$= 60 m$

So gun does the other fragment land at 60m

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First of all maximum height H=u2Sin2θgH=(400×12)2×10H = 10mif they don't explode range OAOA=u2sin2θg=OA=400×sin90∘10=40mTherefore AA=OA2=20mFrom gun it is OB=40+20m =60mSo gun does the other fragment land at 60m