The initial position and velocity of a projectile are given by r₀ = 3i+4j and v₀ = 6i+8j respectively, and where r is in m and v in m/s. The position in the horizontal direction (in m) as a function of time is:

A. x (t) = 6t

B. x (t) = 4 + 8t

C. x (t) = 3 + 6t + The initial position and velocity of a projectile gt²

D. x (t) = 3 + 6t

x (t) = 3 + 6*t

Option D

Explanation:

Given:

- Position vector r_o = 3 i + 4 j

- Initial velocity vector v_o = 6 i + 8 j

Find:

The position in the horizontal direction (in m) as a function of time is:

Solution:

- Using second kinematics equation of motion we have:

x (t) = x_o + v_o, x*t

Where,

x_o : is the initial position in x direction

v_o, x : is the initial velocity in x direction.

From given vectors we have x_o = 3 and v_o, x = 6

Hence, x (t) = 3 + 6*t is the equation of horizontal position.