Two seconds after projection a projectile is travelling in a direction inclined at 30° to the horizontal after one more sec, it is travelling horizontally, the magnitude and direction of its velocity are

The correct answer is 20 √3 m/sec, 60°.

Let us solve the problem step-by-step.

1. Define the Variables:
Let the initial velocity of projection of the projectile be $u$ and its angle of projection with the horizontal be $\theta$.
Let $g$ be the acceleration due to gravity, which we will take as $10{\text{m/s}}^2$.

2. Analyze the Horizontal and Vertical Components of Velocity:
The horizontal component of the velocity remains constant throughout the motion:
$v_x\left(t\right)=ucos\theta$
The vertical component of the velocity at any time $t$ is given by:
$v_y\left(t\right)=usin\theta -gt$

3. Use the Condition at $t=3$ seconds:
According to the problem, after 2 seconds plus one more second (total time $t=3\text{s}$), the projectile is travelling horizontally.
Travelling horizontally means that the vertical component of velocity is zero:
$v_y\left(3\right)=0$
Substituting $t=3$ and $g=10{\text{m/s}}^2$:
$usin\theta -3\left(10\right)=0$
$usin\theta =30\text{m/s}$

4. Use the Condition at $t=2$ seconds:
Two seconds after projection, the direction of motion is inclined at 30° to the horizontal. Therefore:
$tan30°=\frac{v_y\left(2\right)}{v_x\left(2\right)}$
$\frac{1}{\sqrt{3}}=\frac{usin\theta -2g}{ucos\theta }$
Substitute the values of $usin\theta =30$ and $g=10$ into the equation:
$\frac{1}{\sqrt{3}}=\frac{30-2\left(10\right)}{ucos\theta }$
$\frac{1}{\sqrt{3}}=\frac{10}{ucos\theta }$
$ucos\theta =10\sqrt{3}\text{m/s}$

5. Calculate the Angle of Projection ($\theta$):
Dividing the vertical component equation by the horizontal component equation:
$tan\theta =\frac{usin\theta }{ucos\theta }=\frac{30}{10\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}$
Since $tan\theta =\sqrt{3}$, we have:
$\theta =60°$

6. Calculate the Magnitude of Initial Velocity ($u$):
Using the Pythagorean identity:
$u=\sqrt{{(ucos\theta )}^2+{(usin\theta )}^2}$
$u=\sqrt{{\left(10\sqrt{3}\right)}^2+{\left(30\right)}^2}$
$u=\sqrt{300+900}$
$u=\sqrt{1200}=20\sqrt{3}\text{m/s}$

Thus, the magnitude and direction of the velocity of projection are 20 √3 m/sec and 60° respectively.