Question

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

Answer 1

Method 1: Using properties of projectile motion

As we have to calculate the time between two positions A and B where the final direction of movement is perpendicular to the initial direction of movement. So for our own comfortability, we can choose the initial direction of motion as x-axis. Also let us assume the velocity at position B to be $v$.

Now analyzing motion in x-and y-direction, we have

$u_x=u;$ $u_y=0$

$a_x=-g\sin \theta ;a_y=-g\cos \theta$

Here we can use the following formula $v=u+$ at in $x$-direction. As we have the values of initial velocity, final velocity, and acceleration we can find $t$. Therefore,

$v_x=u_x+a_{x}t$

At position $B$, $v_x=0$, as the final velocity is equal to the y-component of velocity. Therefore,

$0=u-g\sin \theta .t$

Thus, $t=\frac{u}{g\sin \theta }$ which is the required time to travel.

Method 2: Using vectors

As $u$ and $v$ both are perpendicular to each other. We can use the orthogonality property of dot product, i.e., if two vectors are perpendicular to each other their dot product is zero, in order to find out the time of travel to the desired position. So,

So, $\overrightarrow{u}.\overrightarrow{v}=0\Rightarrow \overrightarrow{u}.(\overrightarrow{u}+\overrightarrow{a}t)=0\Rightarrow \overrightarrow{u}.\overrightarrow{u}+\overrightarrow{u}.\overrightarrow{a}t=0$

$\Rightarrow u^2+ug\cos ({90}^o+\theta )t=0$

[Because angle between $u$ and $g$ is ${90}^o+\theta$ as from Figure]

$u^2+u.g(-\sin \theta )t=0.t=0$

So, $t=\frac{u}{g\sin \theta }$ is the desired time.

To find out the velocity we can use the same relation as used in this question. But as at final position (considered) only the y-component of velocity is present, so we need to use the same relation in y-direction.

$v_y=u_y+a_{g}t$

$\Rightarrow v=0-g.cos\theta t$

$=-gcos\theta \frac{u}{g\sin \theta }=-u\cot \theta$ is the velocity at position B.

Method 3: If the initial velocity u and velocity at time $t$ are perpendicular, then the final velocity will beat an angle $\theta$ with the vertical.

The horizontal component of velocity is unchanged throughout the motion. Therefore,

$ucos\theta =vsin\theta$

or $v=ucot\theta$

The vertical component of velocity after time $t=-vcos\theta$.

From the equation, $v_y=u\sin \theta -gt-v\cos \theta =u\sin \theta -gt$

$t=\frac{usin\theta +vcos\theta }{g}$

$=\frac{usin\theta +ucot\theta cos\theta }{g}$

$=\frac{u}{g}\left[\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta }\right]=\frac{u}{g}cosec\theta$

Method 4: The slope of trajectory at the point of projection,

$m_1=\tan \theta$.

Slope of trajectory after time $t$,

$m_2=\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{v_y}{v_x}=\frac{u\sin \theta -gt}{u\cos \theta }$

Slopes are perpendicular, $\left(\frac{u\sin \theta -gt}{u\cos \theta }\right)\left(\tan \theta \right)=-1$

or $t=\frac{u}{g\sin \theta }$