A projectile is fired at 30° to the horizontal. The vertical component of its velocity is 80 ms⁻¹ . Its time of flight is T. What will be the velocity of the projectile at t = T/2

The correct answer is 80√3 ms⁻¹.

Let us break down the solution step-by-step:

For a projectile fired with an initial velocity $u$ at an angle $\theta$ to the horizontal, its velocity has two components:
1. A horizontal component $u_x=ucos\theta$, which remains constant throughout the motion because there is no horizontal acceleration (neglecting air resistance).
2. A vertical component $u_y=usin\theta$, which decreases as the projectile goes up due to gravity and increases as it falls down.

We are given:
- The angle of projection, $\theta =30°$
- The initial vertical component of velocity, $u_y=80{\text{ms}}^{-1}$

Using the formula for the vertical component of velocity, we can determine the magnitude of the initial velocity $u$:
$u_y=usin\theta$
$80=usin30°$

Since $sin30°=\frac{1}{2}$:
$80=u\left(\frac{1}{2}\right)\Rightarrow u=160{\text{ms}}^{-1}$

Now, let us calculate the constant horizontal component of the velocity $u_x$:
$u_x=ucos30°$

Substituting $u=160{\text{ms}}^{-1}$ and $cos30°=\frac{\sqrt{3}}{2}$:
$u_x=160\times \frac{\sqrt{3}}{2}$

The total time of flight of the projectile is $T$. The time taken to reach the maximum height is exactly half of the total time of flight, which is:
$t=\frac{T}{2}$

At the maximum height (at $t=\frac{T}{2}$), the vertical component of the velocity becomes zero:
$v_y=0$

Therefore, the net velocity of the projectile at $t=\frac{T}{2}$ consists only of its horizontal component:
$v=v_x=u_x=80\sqrt{3}{\text{ms}}^{-1}$