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: Pick the speed at the top.
At the highest point the up and down speed is zero, so only the sideways speed is left. That sideways speed stays the same the whole flight. Here \[ v = v_0\cos\theta = 40\cos 60^\circ = 40\times \tfrac{1}{2} = 20\text{ m/s} \]

Step 2: Find what pulls it sideways into a curve.
At the top the speed points flat across, and gravity pulls straight down. So gravity is fully sideways to the motion and acts like the turning force. That means the centre-seeking acceleration here is simply $a = g = 10\text{ m/s}^2$.

Step 3: Treat the top of the path as part of a circle.
For circular turning, $a = \dfrac{v^2}{R}$. Rearranging for the radius of curvature: \[ R = \frac{v^2}{a} = \frac{(20)^2}{10} \]

Step 4: Work out the number.
\[ R = \frac{400}{10} = 40\text{ m} \]
So the path bends with a radius of 40 m at the top, which is option (A).
\[ \boxed{40\text{ m}} \]