Question

A particle is projected at angle $\theta$ with horizontal. Calculate the time when it is moving perpendicular to the initial direction.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Let's draw the diagram and see how to proceed.

So it's given that the particle is projected at an angle $\theta$. Let's assume that the velocity of projection is $u$. Now after time $t$ the velocity becomes $v$ and is perpendicular to the direction of projection. By using trigonometry we can find the angle that it makes with x-axis i.e. $(90-\theta )$

So we need to find the time $t$

We know one more thing.

In this particular case $a_x$ = $0$

$\Rightarrow$ The velocity component in the x-direction will be constant.

$\Rightarrow$ $ucos\theta =vcos(90-\theta )$

$\Rightarrow$ $ucos\theta =vsin\theta$

$\Rightarrow$ $v=\frac{ucos\theta }{sin\theta }$ --------------(1)

Now

$u_y=usin\theta$

$v_y=-vsin(90-\theta )=-vcos\theta$

Using equation (1)

$v_y=-\frac{uco{s}^2\theta }{sin\theta }$

$a_y=-g$

Applying equation of Motion
$V_y=u_y+at$
$\frac{-uco{s}^2\theta }{sin\theta }$=$usin\theta -gt$
$\Rightarrow gt=usin\theta +\frac{uco{s}^2\theta }{sin\theta }$
$\Rightarrow gt=u\left[\frac{si{n}^2\theta +co{s}^2\theta }{sin\theta }\right]$
$\Rightarrow t=\frac{u}{gsin\theta }$