A body is projected horizontally from the top of a tower with initial velocity 18 ms⁻¹ . It hits the ground at angle 45°. What is the vertical component of velocity when it strikes the ground

The correct option is 18 ms⁻¹.

Let's break down the motion of the projected body step-by-step:

1. Horizontal Component of Velocity:
When a body is projected horizontally, it undergoes two-dimensional projectile motion. In the absence of air resistance, there is no horizontal force acting on the body. Therefore, the horizontal component of velocity remains constant throughout its flight.
Let $v_x$ be the horizontal component of velocity when the body strikes the ground. We have:
$v_x=u_x=18{\text{ms}}^{-1}$

2. Angle of Impact with the Ground:
The body hits the ground at an angle of $45°$ with the horizontal. Let $\theta$ be this angle. The relation between the vertical component of velocity ($v_y$) and the horizontal component of velocity ($v_x$) at any instant is given by the tangent of the angle of the velocity vector:
$\tan \left(\theta \right)=\frac{v_y}{v_x}$

3. Calculating the Vertical Component of Velocity:
Substituting the given angle $\theta =45°$ into the equation:
$\tan (45°)=\frac{v_y}{v_x}$
Since $\tan (45°)=1$, we get:
$1=\frac{v_y}{v_x}\Rightarrow v_y=v_x$

Substitute the value of $v_x$ into the equation:
$v_y=18{\text{ms}}^{-1}$

Therefore, the vertical component of the velocity when it strikes the ground is 18 ms⁻¹.