Range of a projectile with time of flight 10 s and maximum height 100 m is : `(g= -10 m//s ^2)`

Options

A. 200 m
B. 125 m
C. 100 m
D. Data incoorect

Explanation

To solve the problem of finding the range of a projectile given the time of flight and maximum height, we can follow these steps:

### Step 1: Understand the formulas
We have two key formulas for projectile motion:
1. **Time of Flight (T)**: \( T = \frac{2u \sin \theta}{g} \)
2. **Maximum Height (H)**: \( H = \frac{u^2 \sin^2 \theta}{2g} \)

Where:
- \( u \) = initial velocity
- \( \theta \) = angle of projection
- \( g \) = acceleration due to gravity (given as \( -10 \, \text{m/s}^2 \))

### Step 2: Use the Time of Flight
Given that the time of flight \( T = 10 \, \text{s} \), we can set up the equation:
\[ 10 = \frac{2u \sin \theta}{10} \]
Rearranging gives:
\[ u \sin \theta = 50 \quad \text{(Equation 1)} \]

### Step 3: Use the Maximum Height
Given that the maximum height \( H = 100 \, \text{m} \), we can set up the equation:
\[ 100 = \frac{u^2 \sin^2 \theta}{2 \times 10} \]
This simplifies to:
\[ 100 = \frac{u^2 \sin^2 \theta}{20} \]
Rearranging gives:
\[ u^2 \sin^2 \theta = 2000 \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 1 into Equation 2
From Equation 1, we have \( u \sin \theta = 50 \). Squaring both sides gives:
\[ u^2 \sin^2 \theta = 2500 \quad \text{(Equation 3)} \]

### Step 5: Compare Equations 2 and 3
Now we have:
- From Equation 2: \( u^2 \sin^2 \theta = 2000 \)
- From Equation 3: \( u^2 \sin^2 \theta = 2500 \)

Since both equations cannot be true simultaneously, we conclude that the data provided in the problem is inconsistent.

### Conclusion
Thus, the range cannot be determined from the given data, and the correct conclusion is that the data is incorrect.