Question

A projectile at any instant during its flight has velocity $5\,\,m{s^{ - 1}}$ at ${30^ \circ }$ above the horizontal. How long after this instant, will it be moving at the right angle to the given direction?

Answer 1

Hint: The object that is thrown in a projectile has its own velocity, acceleration and the angle at which it is thrown. Since the time taken is not given by using the given data and one of the three equations of motion, we can solve the given problem.

Useful formula
By using the first equation of motion;
$V{'_y} = {V_y} + at$
Where, $V{'_y}$ denotes the projectile downward velocity, ${V_y}$ denotes the projectiles upward velocity, $a$ can be denoted as the acceleration due to gravity, $t$ denotes as the time taken.

Complete step by step solution
The data given in the problem is;
Velocity of projectile during flight, $V = 5\,m{s^{ - 1}}$ ,
Angle above the horizontal line, $\theta = {30^ \circ }$ .
   

Applying the first equation of motion;
$V{'_y} = {V_y} + at$
For projectile upward velocity ${V_y}$
${V_y} = V \times \sin \theta $
Substitute the values for velocity $V$ and $\theta $ ;
${V_y} = 5\,m{s^{ - 1}}\, \times \sin \left( {{{30}^ \circ }} \right)$
Simplifying the above equation;
${V_y} = 5 \times 0.5$
${V_y} = 2.5\,m{s^{ - 1}}\,\,...........\left( 1 \right)$
For projectile downward velocity $V{'_y}$, $\theta $ is;
$ - \left( {{{90}^ \circ } - {{30}^ \circ }} \right) = - {60^ \circ }$
$V{'_y} = V \times \sin \theta $
Substitute the values of $V$ and $\theta $ ;
$V{'_y} = 5\,m{s^{ - 1}}\, \times \sin \left( { - {{60}^ \circ }} \right)$
Simplifying the above equation;
$V{'_y} = - 5 \times \cos \left( {{{30}^ \circ }} \right)$
$V{'_y} = - 2.5\sqrt 3 \,m{s^{ - 1}}\,\,\,..........\left( 2 \right)$
Substituting the equation (1) and (2) in first equation of motion;
$V{'_y} = {V_y} + at$
Since, $a$is acceleration due to gravity, it can be denoted as $g$;
$V{'_y} = {V_y} + gt$
We know the value of gravity is $g = 9.8\,m{s^{ - 2}}$ we can take it as $g = 10\,m{s^{ - 2}}$
$ - 2.5\sqrt 3 \,m{s^{ - 1}}\, = 2.5\,m{s^{ - 1}}\, + 10\,m{s^{ - 2}}\, \times t$

Since we only need the time taken $t$ ;
$t = \dfrac{{ - 2.5\sqrt 3 - 2.5}}{{ - 10}}$
$t = 0.68\,s$

Therefore, after $t = 0.68\,s$ it will be moving at right angle to the given direction.

Note: There are three types of equations of motion. The first equation of motion deals with the initial velocity and time. The second equation of motion deals with the distance, time initial velocity and acceleration’s relationship. The third equation of motion deals with the distance, final velocity, acceleration and initial velocity.