A projectile is fired with velocity u making angle θ with the horizontal. What is the change in velocity when it is at the highest point

To find the change in velocity of the projectile when it reaches its highest point, we analyze the velocity components at the point of projection and at the highest point.

Let the horizontal direction be the positive x-axis and the vertical direction be the positive y-axis.

At the point of projection, the projectile is fired with an initial velocity u at an angle $\theta$ with the horizontal. The initial velocity vector ${\overrightarrow{v}}_i$ can be written in terms of its horizontal and vertical components:
${\overrightarrow{v}}_i=\left(u\cos \theta \right)\hat{i}+\left(u\sin \theta \right)\hat{j}$

During projectile motion, there is no acceleration in the horizontal direction (neglecting air resistance). Therefore, the horizontal component of the velocity remains constant throughout the motion:
$v_x=u\cos \theta$

At the highest point of the trajectory, the vertical component of the velocity becomes zero:
$v_y=0$

Thus, the velocity vector at the highest point ${\overrightarrow{v}}_f$ is:
${\overrightarrow{v}}_f=\left(u\cos \theta \right)\hat{i}+0\hat{j}=\left(u\cos \theta \right)\hat{i}$

The change in velocity vector $\Delta \overrightarrow{v}$ is the difference between the final velocity and the initial velocity:
$\Delta \overrightarrow{v}={\overrightarrow{v}}_f-{\overrightarrow{v}}_i$
$\Delta \overrightarrow{v}=\left[\right(u\cos \theta \left)\hat{i}\right]-\left[\right(u\cos \theta )\hat{i}+(u\sin \theta \left)\hat{j}\right]$
$\Delta \overrightarrow{v}=-\left(u\sin \theta \right)\hat{j}$

The magnitude of this change in velocity is:
$\left|\Delta \overrightarrow{v}\right|=u\sin \theta$

Therefore, the magnitude of the change in velocity when the projectile is at the highest point is u sinθ.