Question

A body is projected horizontally from the top of a building. It strikes the ground after a time $t$ with its velocity vector making an angle $\theta$ with the horizontal. The speed with which the body is projected is

• A. $\frac{gt}{\sin \theta }$
• B. $\frac{gt}{\cos \theta }$
• C. $\frac{gt}{\tan \theta }$
• D. $\frac{gt}{\cot \theta }$

Answer 1

The correct option is C $\frac{gt}{\tan \theta }$

Let $u$ be the speed with which the body is projected and let $v_x$ and $v_y$ be the horizontal and vertical components of the velocity at time t. Then
$\tan \theta =\frac{v_y}{v_x}=\frac{v_y}{u}(\because v_x=u$ remains constant)
$\Rightarrow u=\frac{v_y}{\tan \theta }\left(i\right)$
Now, since the initial y component $\left(u_y\right)$ is zero, we have
$-v_y=0-gt$
$\Rightarrow v_y=gt\left(ii\right)$
Using (ii) in (i), we get
$u=\frac{gt}{\tan \theta }$, which is choice (c)