A stone is projected in air. Its time of flight is 3s and range is 150m. Maximum height reached by the stone is (Take, `g = 10 ms^(-2))`

Options

A. `37.5m`
B. `22.5 m`
C. `90 m`
D. `11.25m`

Explanation

To solve the problem, we need to find the maximum height reached by a stone projected in the air, given its time of flight and range. We will follow these steps:

### Step 1: Understand the given data
- Time of flight (T) = 3 seconds
- Range (R) = 150 meters
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Use the time of flight formula
The time of flight for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
where \( u \) is the initial velocity and \( \theta \) is the angle of projection.

For maximum height, we consider the angle \( \theta = 90^\circ \) (which is not the case here, but we will derive \( u \) first).

Rearranging the formula for \( u \):
\[ u = \frac{gT}{2 \sin \theta} \]

### Step 3: Calculate the initial velocity (u)
Since we don't have the angle \( \theta \), we can use the range formula to find \( u \) first. The range \( R \) is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]

### Step 4: Use the range formula
We can express \( \sin 2\theta \) as \( 2 \sin \theta \cos \theta \):
\[ R = \frac{u^2 (2 \sin \theta \cos \theta)}{g} \]

### Step 5: Substitute the known values
Substituting \( R = 150 \) m and \( g = 10 \) m/s²:
\[ 150 = \frac{u^2 (2 \sin \theta \cos \theta)}{10} \]
\[ 1500 = u^2 (2 \sin \theta \cos \theta) \]

### Step 6: Use the time of flight to find \( u \)
From the time of flight equation:
\[ 3 = \frac{2u \sin \theta}{10} \]
\[ 30 = 2u \sin \theta \]
\[ u \sin \theta = 15 \]

### Step 7: Substitute \( u \sin \theta \) into the range equation
Now we can substitute \( u \sin \theta = 15 \) into the range equation:
\[ 1500 = u^2 (2 \cdot \frac{15}{u}) \]
\[ 1500 = 30u \]
\[ u = \frac{1500}{30} = 50 \text{ m/s} \]

### Step 8: Calculate the maximum height
The formula for maximum height \( h \) is:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]

Using \( u = 50 \) m/s and substituting \( \sin \theta \):
From \( u \sin \theta = 15 \):
\[ \sin \theta = \frac{15}{50} = 0.3 \]

Now substituting into the height formula:
\[ h = \frac{50^2 \cdot (0.3)^2}{2 \cdot 10} \]
\[ h = \frac{2500 \cdot 0.09}{20} \]
\[ h = \frac{225}{20} = 11.25 \text{ meters} \]

### Final Answer
The maximum height reached by the stone is **11.25 meters**.

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