Question

a projectile is fired at a speed of hundred metre per second at an angle of 37 degrees above the horizontal .At the highest point the projectile breaks into two parts of mass ratio 1:3 the smaller coming to rest .find the distance from the launching point where the heavier piece lands​

Answer 1

• Distance from the launching point where the heavier piece lands is 1120 m.

Given-

Angle = 37 degree

Mass ratio = 1:3

So, velocity of the projectile at the highest point is = 100 (Cos 37) = 80 m/sec

Let the mass of the projectile be 4m.

And mass of the heavier particle be 3m.

Let the velocity of 3m after collision be v.

Therefore by applying conservation of momentum we get-

3mv = 4m (80) = 320m

v = 320/3 m/s

Let the height at which the projectile break be h.

Therefore, h = (100 sin37)^2/2g = 180 m

Time taken by 3m to reach ground =

$\sqrt \frac{2h}{g}$ = $\sqrt{36}$ = 6 sec

Hence by putting the value we get-

Horizontal distance travelled by 3m = (320/3) x 6 = 640 m

Horizontal distance travelled by 4m = 80 x 6 = 480 m

So, Total distance travelled = 640 + 480 = 1120 m

Regards