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. 10 m

Explanation

To solve the problem step by step, we will use the equations of motion for projectile motion.

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

### Step 2: Substitute the given values
We know from the problem that the time of flight \( T = 4 \) seconds and \( g = 10 \, \text{m/s}^2 \). Plugging these values into the time of flight equation:
\[ 4 = \frac{2u \sin \theta}{10} \]

### Step 3: Rearranging the equation to find \( u \sin \theta \)
To isolate \( u \sin \theta \), we can rearrange the equation:
\[ 2u \sin \theta = 4 \times 10 \]
\[ 2u \sin \theta = 40 \]
\[ u \sin \theta = 20 \quad \text{(Equation 1)} \]

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

### Step 5: Substitute \( u \sin \theta \) into the height formula
We can express \( u^2 \sin^2 \theta \) in terms of \( (u \sin \theta)^2 \):
\[ u^2 \sin^2 \theta = (u \sin \theta)^2 \]
Substituting \( u \sin \theta = 20 \) from Equation 1:
\[ h = \frac{(20)^2}{2g} \]
\[ h = \frac{400}{2 \times 10} \]
\[ h = \frac{400}{20} \]
\[ h = 20 \, \text{meters} \]

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

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