A projectile is projected with initial velocity (6î+8ĵ)m/sec.If g=10 ms⁻², then horizontal range is

The correct option is 9.6 metre.

Step-by-Step Explanation:

1. Identify the given components of velocity:
The initial velocity vector is given as:
$u=6\hat{i}+8\hat{j}{\text{ms}}^{-1}$
From this, we can extract the horizontal component ($u_x$) and the vertical component ($u_y$) of the initial velocity:
Horizontal component of velocity, $u_x=6{\text{ms}}^{-1}$
Vertical component of velocity, $u_y=8{\text{ms}}^{-1}$
Acceleration due to gravity, $g=10{\text{ms}}^{-2}$

2. Formula for Horizontal Range:
The horizontal range ($R$) of a projectile is the total horizontal distance covered during its flight. It can be expressed as:
$R=u_x\times T$
where $T$ is the total time of flight.

3. Calculate the Time of Flight ($T$):
The time of flight is determined by the vertical motion of the projectile and is given by:
$T=\frac{2u_y}{g}$
Substituting the given values:
$T=\frac{2\times 8}{10}=\frac{16}{10}=1.6\text{seconds}$

4. Calculate the Horizontal Range ($R$):
Now, substitute the values of $u_x$ and $T$ back into the range formula:
$R=6\times 1.6=9.6\text{metres}$

Thus, the horizontal range of the projectile is 9.6 metre.