Question

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 _______.

• A. 60
• B. 45
• C. 30
• D. 75

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The velocity vector components of a projectile can be found by taking the time derivative of its position coordinates. The angle of the projectile's trajectory at any given time is the angle of its velocity vector relative to the horizontal plane.

Step 2: Key Formula or Approach:
Velocity components: $v_x = \frac{dx}{dt}$ and $v_y = \frac{dy}{dt}$.
Angle with the horizontal $\theta$: $\tan \theta = \frac{v_y}{v_x}$.

Step 3: Detailed Explanation:
Given the position coordinates:
$x = 24t$
$y = 43.6t - 4.9t^2$
Differentiate to find velocity components:
$v_x = \frac{d}{dt}(24t) = 24 \text{ m/s}$ (Constant horizontal velocity)
$v_y = \frac{d}{dt}(43.6t - 4.9t^2) = 43.6 - 9.8t \text{ m/s}$
Evaluate the velocity components at the specific time $t = 2$ s:
$v_x = 24 \text{ m/s}$
$v_y = 43.6 - 9.8(2) = 43.6 - 19.6 = 24 \text{ m/s}$
The angle $\theta$ made with the horizontal is given by:
$\tan \theta = \frac{v_y}{v_x} = \frac{24}{24} = 1$
Since $\tan \theta = 1$ and both components are positive, the angle is in the first quadrant:
$\theta = 45^\circ$.

Step 4: Final Answer:
The angle is $45^\circ$.