A projectile fired from the ground follows a parabolic path. The speed of the projectile is minimum at the top of its path.

Step 1: Define the initial parameters of the projectile.

Let a projectile be fired from the ground with an initial speed $u$ at an angle $\theta$ to the horizontal.

$$\vec{u} = u\cos\theta\, \hat{i} + u\sin\theta\, \hat{j}$$

Step 2: Determine the velocity components at any time $t$.

The horizontal velocity remains constant due to the absence of horizontal forces, while the vertical velocity is affected by gravity.

$$v_x = u\cos\theta$$ $$v_y = u\sin\theta - gt$$

Step 3: Express the speed of the projectile as a function of time.

The speed $v(t)$ is the magnitude of the velocity vector.

$$v(t) = \sqrt{v_x^{2} + v_y^{2}}$$ $$v(t) = \sqrt{(u\cos\theta)^{2} + (u\sin\theta - gt)^{2}}$$

Step 4: Find the velocity components at the top of the path.

At the highest point of the parabolic path, the vertical component of velocity becomes zero.

$$v_y = 0$$ $$t = \frac{u\sin\theta}{g}$$

Step 5: Calculate the minimum speed.

Since the squared term $(u\sin\theta - gt)^{2} \ge 0$, the speed is minimized when this term is zero.

$$v_{\text{min}} = \sqrt{(u\cos\theta)^{2} + 0}$$ $$v_{\text{min}} = u\cos\theta$$

This minimum speed occurs at the top of the path.

Answer: The statement is TRUE. The speed of the projectile is minimum at the top of its path, with a value of $u\cos\theta$.