If \(x\) and \(y\) coordinates of a projectile as a function of time \((t)\) are given as \(24t\) and \(43.6t-4.9t^2\), respectively, then the angle (in degrees) made by the projectile with horizontal when \(t=2\) s is ______.

The velocity components of the projectile are given by the derivatives of the position coordinates with respect to time.

\(v_x = \dfrac{dx}{dt} = \dfrac{d}{dt}(24t) = 24\)

\(v_y = \dfrac{dy}{dt} = \dfrac{d}{dt}(43.6t - 4.9t^2) = 43.6 - 9.8t\)

At \(t = 2\) s, the velocity components are:

\(v_x = 24\)

\(v_y = 43.6 - 9.8(2) = 43.6 - 19.6 = 24\)

The angle \(\theta\) made by the velocity vector with the horizontal is given by:

\(\tan \theta = \dfrac{v_y}{v_x} = \dfrac{24}{24} = 1\)

\(\theta = 45^{\circ}\)

Answer: \(45\)