Q54 / 75 · JEE Main 2026 (04 April Shift 2) · Physics
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$60$
- Correct answerB$45$
- C$30$
- D$75$
Solution
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: $45$
$v_x = \dfrac{dx}{dt} = \dfrac{d}{dt}(24t) = 24$
$v_y = \dfrac{dy}{dt} = \dfrac{d}{dt}(43.6t - 4.9t^2) = 43.6 - 9.8t$
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: $45$
Hence, the correct answer is option: 'B'.