Question

A body is projected horizontally from the top of a tower with initial velocity $18m{s}^{-1}$. It hits the ground at angle ${45}^{\circ }$. What is the vertical component of velocity when strikes the ground?

• A. $18\sqrt{2}m{s}^{-1}$
• B. $18m{s}^{-1}$
• C. $9\sqrt{2}m{s}^{-1}$
• D. $9m{s}^{-1}$

Answer 1

The correct option is B $18m{s}^{-1}$

$v\cos {45}^{\circ }=u=18m{s}^{-1}\Rightarrow v=18\sqrt{2}m{s}^{-1}$
Vertical component $v\sin {45}^{\circ }=18\sqrt{2}\times \frac{1}{\sqrt{2}}=18m{s}^{-1}$.