Question

A projectile explodes at its highest point into 3 equal parts. One part retraces its path back, the second part
loses its velocity and the third part lands on the horizontal surface (the surface from where projectile was
initially projected) at the distance of 30 root 3 m from the point at which original projectile would have landed at
the horizontal surface. If the projectile was projected at an angle of 30°, find the velocity of projection (in m/s):
a) 10
b) 20
c) 10 root 2
d) 20 root 2
Can anyone EXPLAIN this answer please?? I need it now ​

Answer 1

Given :

Angle of projection = 30°

After explosion,

• First particle come back to the initial point
• Second particle lands at midpoint
• Third particle lands at a distance of 30√3 m from the point P (COM)

To Find :

Velocity of projectile.

Solution :

In projectile motion, no horizontal force acts on projectile body hence centre of mass won't change its path after explosion.

Assume that, P point is the COM!

• Distance of first particle from COM = -R (equal to range of projectile)
• Distance of second particle from COM = -R/2 (midpoint or half of total range)
• Distance of third particle from COM = 30√3 m (given in the question)

Let coordinates of point P be (0,0) and mass of each particle be m.

From the definition of COM :-

$\sf:\implies\:x=\dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}$

$\sf:\implies\:0=\dfrac{(m)(-R)+(m)(-R/2)+m(30\sqrt3)}{m+m+m}$

$\sf:\implies\:0=\dfrac{-3R/2+30\sqrt3}{3}$

$\sf:\implies\:3R/2=30\sqrt3$

$\bf:\implies\:R=20\sqrt3\:m$

★ Range of projectile :

$\sf:\implies\:R=\dfrac{u^2sin2\theta}{g}$

$\sf:\implies\:20\sqrt3=\dfrac{u^2sin2(30^{\circ})}{10}$

$\sf:\implies\:200\sqrt3=u^2sin60^{\circ}$

$\sf:\implies\:200\sqrt3=u^2\times \dfrac{\sqrt3}{2}$

$\sf:\implies\:u^2=400$

$\bf:\implies\:\boxed{\bf{u=20\:ms^{-1}}}$

∴ (B) is the correct answer!

Answer 2

Given :

Angle of projection = 30°

After explosion,

First particle come back to the initial point

Second particle lands at midpoint

Third particle lands at a distance of 30√3 m from the point P (COM)

To Find :

Velocity of projectile.

Solution :

In projectile motion, no horizontal force acts on projectile body hence centre of mass won't change its path after explosion.

Assume that, P point is the COM!

Distance of first particle from COM = -R (equal to range of projectile)

Distance of second particle from COM = -R/2 (midpoint or half of total range)

Distance of third particle from COM = 30√3 m (given in the question)

Let coordinates of point P be (0,0) and mass of each particle be m.

From the definition of COM :-