Question

A projectlie is fired from a gun at an angle of 45° with the horizantal and with speed 20m/s relative to ground. At the highest point in its flight the projectile explodes into two fragments of equal mass. One fragment comes rest just after explosion. How far from the gun does the other fragment land, assuming a horizontal ground. g=10m/s²

Answer 1

Refer to the attachments

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

First of all maximum height,

H = u²sin²θ / g

H = (400 × 1/2) /2 × 10

H = 10 m.

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

If they don't break (explode) range  OA

OA = u² sin2θ /g

OA = 400 × sin 45×2 /10

OA = 400 × sin 90° /10

OA = 40 m.

Therefore ,

AA' = OA/2

AA' = 20 m.

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 +20 m

= 60 m.

So gun does the other fragment land at 60 m.

Hope it helps.°