A body is projected with an angle `theta`. The maximum height reached is h. If the time of flight is 4 sec and `g=10m//s^2`, then the value of h is 

Options

A. 40 m
B. 20 m
C. 5 m
D. 10m

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between time of flight, initial velocity, and angle of projection.
The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2U \sin \theta}{g} \]
where \( U \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

### Step 2: Substitute the known values into the time of flight equation.
Given that the time of flight \( T = 4 \) seconds and \( g = 10 \, \text{m/s}^2 \):
\[ 4 = \frac{2U \sin \theta}{10} \]

### Step 3: Rearrange the equation to solve for \( U \sin \theta \).
Multiplying both sides by \( 10 \):
\[ 40 = 2U \sin \theta \]
Now, divide both sides by 2:
\[ U \sin \theta = 20 \]

### Step 4: Use the formula for maximum height.
The maximum height \( h \) reached by the projectile is given by:
\[ h = \frac{U^2 \sin^2 \theta}{2g} \]

### Step 5: Substitute \( U \sin \theta \) into the height formula.
From the previous step, we have \( U \sin \theta = 20 \). Therefore, we can express \( U \) in terms of \( \sin \theta \):
\[ U = \frac{20}{\sin \theta} \]

### Step 6: Substitute \( U \) back into the height formula.
Now substituting \( U \) into the height formula:
\[ h = \frac{\left(\frac{20}{\sin \theta}\right)^2 \sin^2 \theta}{2g} \]
This simplifies to:
\[ h = \frac{400}{\sin^2 \theta} \cdot \frac{\sin^2 \theta}{2g} \]
\[ h = \frac{400}{2g} \]

### Step 7: Substitute the value of \( g \).
Now substitute \( g = 10 \, \text{m/s}^2 \):
\[ h = \frac{400}{2 \times 10} = \frac{400}{20} = 20 \, \text{m} \]

### Final Answer:
The maximum height \( h \) reached by the body is \( 20 \, \text{m} \).

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