A body is thrown at angle 30° to the horizontal with the velocity of 30 m/s. After 1 sec, its velocity will be (in m/s) (g = 10 m/s²)

The correct option is 10 √7.

To find the velocity of the body after 1 second, we need to analyze its motion in both the horizontal and vertical directions separately. This is a classic projectile motion problem.

First, let us write down the given values:
Initial velocity, $u=30\text{m/s}$
Angle of projection, $\theta ={30}^{���}$
Time, $t=1\text{s}$
Acceleration due to gravity, $g=10{\text{m/s}}^2$

Step 1: Calculate the horizontal component of velocity ($v_x$)
Since there is no horizontal acceleration in projectile motion (neglecting air resistance), the horizontal component of velocity remains constant throughout the motion.
$v_x=ucos\theta$
Substituting the given values:
$v_x=30cos\left({30}^{\circ }\right)=30\times \frac{\sqrt{3}}{2}=15\sqrt{3}\text{m/s}$

Step 2: Calculate the vertical component of velocity ($v_y$)
The vertical velocity changes over time due to the acceleration of gravity acting downwards. Using the first equation of motion for the vertical direction:
$v_y=usin\theta -gt$
Substituting the given values:
$v_y=30sin\left({30}^{\circ }\right)-10\times 1$
$v_y=30\times \frac{1}{2}-10=15-10=5\text{m/s}$

Step 3: Calculate the net velocity ($v$) after 1 second
The net velocity is the vector sum of its horizontal and vertical components:
$v=\sqrt{v_x^2+v_y^2}$
Substituting the values of $v_x$ and $v_y$:
$v=\sqrt{{\left(15\sqrt{3}\right)}^2+5^2}$
$v=\sqrt{225\times 3+25}$
$v=\sqrt{675+25}=\sqrt{700}$
$v=10\sqrt{7}\text{m/s}$

Therefore, the velocity of the body after 1 second is 10 √7 m/s.