Question

Two identical bodies, projected with the same speed at two different angles cover the same horizontal range \(R\). If the time of flight of these bodies are 5 s and 10 s, respectively, then the value of \(R\) is ________ m. (Take \(g = 10\) m/s²)

• A. 250
• B. 25
• C. 500
• D. 125

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a projectile, if two different angles of projection give the same horizontal range for a given initial speed, the angles must be complementary, i.e., $\theta$ and $90^\circ - \theta$. There is a direct mathematical relationship between their respective times of flight and the horizontal range.
Step 2: Key Formula or Approach:
1. Time of flight: $t = \frac{2u \sin\theta}{g}$.
2. Horizontal Range: $R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin\theta \cos\theta}{g}$.
3. Relationship: $t_1 t_2 = \frac{2R}{g}$.
Step 3: Detailed Explanation:
Let the initial speed be $u$. Since ranges are equal, the angles are $\theta$ and $90^\circ - \theta$.
Time of flight for the first body:
$t_1 = \frac{2u \sin\theta}{g} = 5\text{ s}$.
Time of flight for the second body:
$t_2 = \frac{2u \sin(90^\circ - \theta)}{g} = \frac{2u \cos\theta}{g} = 10\text{ s}$.
Multiply $t_1$ and $t_2$:
$t_1 \times t_2 = \left( \frac{2u \sin\theta}{g} \right) \times \left( \frac{2u \cos\theta}{g} \right) = \frac{4u^2 \sin\theta \cos\theta}{g^2}$.
We can rearrange this product to match the formula for Range $R$:
$t_1 t_2 = \frac{2}{g} \left( \frac{2u^2 \sin\theta \cos\theta}{g} \right) = \frac{2}{g} \left( \frac{u^2 \sin 2\theta}{g} \right) = \frac{2R}{g}$.
Now, substitute the known values:
$5 \times 10 = \frac{2R}{10}$.
$50 = \frac{R}{5}$.
$R = 50 \times 5 = 250\text{ m}$.
Step 4: Final Answer:
The value of R is $250\text{ m}$.