Angle Made by Projectile with Horizontal at Given Time

To find the angle made by the velocity vector of the projectile with the horizontal at $t = 2\text{ s}$, we first calculate the horizontal and vertical components of the velocity as functions of time $t$.

The position coordinates of the projectile are given as: $x(t) = 24t$ $y(t) = 43.6t - 4.9t^2$

Differentiating $x(t)$ and $y(t)$ with respect to time $t$, we obtain the velocity components:

1. Horizontal velocity component ($v_x$): $v_x = \frac{dx}{dt} = 24\text{ m/s}$

2. Vertical velocity component ($v_y$): $v_y = \frac{dy}{dt} = 43.6 - 9.8t\text{ m/s}$

At $t = 2\text{ s}$, substituting $t = 2$ into the velocity expressions yields: $v_x = 24\text{ m/s}$ $v_y = 43.6 - 9.8(2) = 43.6 - 19.6 = 24\text{ m/s}$

The angle $\theta$ made by the projectile's velocity vector with the horizontal is given by: $\tan\theta = \frac{v_y}{v_x}$

Substituting the values of $v_x$ and $v_y$: $\tan\theta = \frac{24}{24} = 1$

$\theta = \tan^{-1}(1) = 45^\circ$

Thus, the angle made by the projectile with the horizontal at $t = 2\text{ s}$ is $45^\circ$.