A projectile is thrown from a point in a horizontal place such that its horizontal and vertical velocity component are 9.8 m/s and 19.6 m/s respectively. Its horizontal range is

The correct option is 39.2 m.

Step-by-step Explanation:

We are given the components of the initial velocity of a projectile:
Horizontal component of velocity, $u_x=9.8\text{m/s}$
Vertical component of velocity, $u_y=19.6\text{m/s}$
Acceleration due to gravity is $g=9.8\text{m/s}{²}^{}$ .

1. Find the Time of Flight (T):
The time of flight depends entirely on the vertical component of the velocity and gravity. It is given by the formula:
$T=\frac{2u_y}{g}$
Substituting the given values:
$T=\frac{2\times 19.6}{9.8}$
$T=2\times 2=4\text{s}$

2. Calculate the Horizontal Range (R):
The horizontal range is the distance traveled horizontally during the time of flight. Since there is no acceleration in the horizontal direction, the range is simply the product of the horizontal velocity component and the time of flight:
$R=u_x\times T$
Substituting the values:
$R=9.8\times 4$
$R=39.2\text{m}$

Therefore, the horizontal range of the projectile is 39.2 m.