Question

A projectile is thrown into space so that it attains a maximum possible range of $200\text{ m}$. Taking the point of projection as origin, the coordinates of the point where the velocity of projectile is minimum is

• A. $(100,50)$
• B. $(200, 100)$
• C. $(100, 200)$
• D. $(50,100)$

Hint:

For a maximum range launch angle ($\theta = 45^\circ$), the maximum peak height achieved is always exactly one-quarter of the total range ($H = R/4$).

Answer 1

The Correct Option is A

Step 1: Concept
The horizontal range of a projectile is maximized when the projection angle is $\theta = 45^\circ$, where $R_{\max} = \frac{u^2}{g}$. The velocity of a projectile is at its minimum at the highest point of its trajectory, where the vertical component of velocity completely vanishes.

Step 2: Meaning
We are given $R_{\max} = 200\text{ m}$. The coordinates of the highest point of any standard trajectory are given by $(x, y) = \left(\frac{R}{2}, H\right)$, where $R$ is the range and $H$ is the maximum height.

Step 3: Analysis
Since the launch angle for maximum range is $45^\circ$, the maximum height formula simplifies beautifully: $H = \frac{u^2\sin^2(45^\circ)}{2g} = \frac{u^2(1/2)}{2g} = \frac{u^2}{4g} = \frac{R_{\max}}{4}$. Given $R_{\max} = 200\text{ m}$, we find the height value: $H = \frac{200}{4} = 50\text{ m}$.

Step 4: Conclusion
The x-coordinate is half of the total range: $x = \frac{200}{2} = 100\text{ m}$. Therefore, the coordinates of this minimum velocity apex point are $(100, 50)$, matching option (A).

Final Answer: (A)