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: Know when range is maximum.
A projectile travels the farthest when it is thrown at $45^{\circ}$. At that angle the range is the biggest possible.

Step 2: Know where speed is least.
At the highest point the up-down part of the velocity becomes zero, so only the sideways part is left. That is where the speed is smallest.

Step 3: Find the x of the top point.
The path is symmetric, so the highest point is at the middle of the range. \[ x = \frac{R}{2} = \frac{200}{2} = 100 \text{ m} \]
Step 4: Use the height for a $45^{\circ}$ throw.
For maximum range the height comes out as $H = \dfrac{R}{4}$. This is a handy result for the $45^{\circ}$ case.

Step 5: Put in the numbers.
\[ H = \frac{200}{4} = 50 \text{ m} \]
Step 6: Write the point.
So the lowest-speed point is at $(100, 50)$, which is option 1.
\[ \boxed{(100,\,50)} \]