Question

A projectile is thrown with a speed of $40\text{ m/s}$ at an angle of $60^\circ$ with the horizontal. Its radius of curvature at the highest point of its trajectory is ($g = 10\text{ m/s}^2$):

• A. 40 m
• B. 20 m
• C. 80 m
• D. 10 m Correct Answer: (A) 40 m

Hint:

Remember that the radius of curvature is at its absolute minimum at the peak of a projectile's trajectory because the path is bending most sharply there. The neat simplified formula $R = \frac{v_h^2}{g}$ acts as a super-fast shortcut for apex evaluation on competitive exams!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A projectile moves along a curved parabolic trajectory under the constant downward pull of gravitational acceleration. The radius of curvature at any given point on a curved path characterizes how sharply the path bends. Locally, we can treat that segment of the path as a circle. The acceleration component acting perpendicular to the direction of motion serves as the centripetal acceleration ($a_c$) keeping the body on this curved arc. At the peak or highest point of the trajectory, the velocity vector is completely horizontal, and the gravitational acceleration points straight down, making them perfectly perpendicular.

Step 2: Key Formula or Approach:
The general relationship for centripetal acceleration in curvilinear motion is: $$ a_n = \frac{v^2}{R} \implies R = \frac{v^2}{a_n} $$ Where: - $R$ is the local radius of curvature. - $v$ is the instantaneous magnitude of velocity at that point. - $a_n$ is the normal (perpendicular) component of acceleration. At the highest point of projectile motion: 1. The vertical component of velocity becomes zero ($v_y = 0$). Only the constant horizontal component remains active: $$ v = v_h = v_0 \cos\theta $$ 2. The total acceleration acting on the projectile is gravity ($g$), directed vertically downward. Since the velocity is strictly horizontal at the apex, gravity acts entirely normal to the velocity vector: $$ a_n = g $$ Therefore, the radius of curvature formula at the peak simplifies to: $$ R = \frac{(v_0 \cos\theta)^2}{g} $$

Step 3: Detailed Explanation:
Let's substitute the given values into our derived relation: - Initial speed $v_0 = 40\text{ m/s}$ - Angle of projection $\theta = 60^\circ$ - Acceleration due to gravity $g = 10\text{ m/s}^2$ 1. Calculate the horizontal velocity component ($v_h$): $$ v_h = 40 \times \cos(60^\circ) $$ Knowing that $\cos(60^\circ) = \frac{1}{2}$: $$ v_h = 40 \times \frac{1}{2} = 20\text{ m/s} $$ 2. Calculate the radius of curvature ($R$) using the centripetal relation: $$ R = \frac{v_h^2}{g} = \frac{(20)^2}{10} $$ $$ R = \frac{400}{10} = 40\text{ m} $$ This result corresponds exactly to option (A).

Step 4: Final Answer:
The radius of curvature of the projectile at the highest point of its trajectory is $40\text{ m}$.