1. Projectile motion basics

For a body projected from the ground with initial speed $u$ at angle $\theta$ with the horizontal:

• Horizontal velocity component: $u_x = u\cos\theta$
• Vertical velocity component: $u_y = u\sin\theta$

During motion under gravity:

• Horizontal velocity remains constant: $v_x = u\cos\theta$
• Vertical velocity changes as: $v_y = u\sin\theta - gt$

At the highest point of projectile motion:

$v_y = 0$

but

$v_x = u\cos\theta \neq 0$

(assuming the projectile was actually projected at a non-vertical angle).

2. Check each option

Option A: Gravitational potential energy is maximum at the highest point.

Gravitational potential energy is

$U = mgh$

It depends on height $h$. Since the height is greatest at the highest point, $U$ is maximum there.

✅ Option A is correct.

Option B: The vertical component of momentum is maximum at the highest point.

Vertical momentum is

$p_y = mv_y$

At the highest point, $v_y = 0$, so

$p_y = 0$

Hence it is not maximum there.

❌ Option B is incorrect.

Option C: The horizontal component of velocity is zero at the highest point.

In projectile motion (neglecting air resistance), horizontal velocity remains constant:

$v_x = u\cos\theta$

So at the highest point it is still non-zero.

❌ Option C is incorrect.

Option D: The Kinetic Energy (K.E.) is zero at the highest point of projectile motion.

Kinetic energy is

$K = \frac{1}{2}mv^2$

At the highest point, although $v_y = 0$, the horizontal component $v_x = u\cos\theta$ still exists. Therefore speed is not zero:

$v = v_x = u\cos\theta$

So kinetic energy is

$K = \frac{1}{2}m(u\cos\theta)^2 \neq 0$

❌ Option D is incorrect.

3. Final conclusion

The only correct statement is:

$\boxed{\text{A}}$

4. Comparison with stored correct answer

Stored correct answer: $\text{A}$

My derived answer: $\text{A}$

They match.