Physics - Kinematics Question with Solution | TestHub

The horizontal velocity of a projectile remains constant throughout its flight, assuming no air resistance. The average velocity between any two points is the displacement divided by the time taken.

Let $H_{max}$ be the maximum height. The projectile crosses half the maximum height ($H_{max}/2$) twice: once on the way up and once on the way down.

The horizontal velocity component is $v_x=u\cos \theta$.

The vertical velocity component is $v_y=u\sin \theta$.

At any point in the projectile's trajectory, the horizontal velocity is constant.

The average velocity is given by ${\vec{v}}_{avg}=\frac{\Delta \vec{r}}{\Delta t}$.

Since the initial and final horizontal positions are $x_1$ and $x_2$, and the initial and final vertical positions are $y_1$ and $y_2$, the displacement is $\Delta \vec{r}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}$.

When the projectile crosses half the maximum height, say at times $t_1$ (ascending) and $t_2$ (descending), the vertical displacement $\Delta y=y_2-y_1=0$ because it returns to the same height.

Therefore, the average vertical velocity between these two points is zero.

The average horizontal velocity is $\frac{x_2-x_1}{t_2-t_1}$.

Since the horizontal velocity $v_x=u\cos \theta$ is constant, the average horizontal velocity is also $u\cos \theta$.

Thus, the average velocity vector is ${\vec{v}}_{avg}=(u\cos \theta )\hat{i}$.

The magnitude of the average velocity is $u\cos \theta$.

The final answer is $\boxed{u\cos \theta }$.