A projectile is fired at 30° with momentum p. Neglecting friction, the change in kinetic energy when it returns to the ground will be

To determine the change in kinetic energy when a projectile returns to the ground, we can analyze the motion of the projectile under the influence of gravity alone, neglecting air resistance and friction.

Let the projectile be launched from the ground with an initial velocity $v$ at a projection angle of 30° with the horizontal.
The initial kinetic energy ($K_i$) of the projectile is given by:
$K_i=\frac{1}{2}m{v}^2$
where $m$ is the mass of the projectile.

In the absence of air resistance, the horizontal component of velocity remains constant throughout the flight:
$v_{x,f}=v_{x,i}=v\cos (30°)$

By symmetry, when the projectile returns to the same vertical level (the ground) from which it was launched, the vertical component of its velocity has the same magnitude but points in the opposite direction:
$v_{y,f}=-v_{y,i}=-v\sin (30°)$

The final speed ($v_f$) when it hits the ground is:
$v_f=\sqrt{{v_{x,f}}^2+{v_{y,f}}^2}=\sqrt{{\left(v\cos \right(30°\left)\right)}^2+{(-v\sin (30°\left)\right)}^2}=v$

Therefore, the final kinetic energy ($K_f$) upon returning to the ground is:
$K_f=\frac{1}{2}m{v_f}^2=\frac{1}{2}m{v}^2=K_i$

The change in kinetic energy ($\Delta K$) is:
$\Delta K=K_f-K_i=0$

Alternatively, by the law of conservation of mechanical energy, since there is no air resistance (friction), the total mechanical energy (potential energy + kinetic energy) is conserved. Since the projectile returns to the same height (the ground), its potential energy is unchanged, which means its kinetic energy must also remain unchanged.

Thus, the change in kinetic energy is zero.

The correct option is Zero.