A body is thrown with some velocity from the ground. Maximum height when it is thrown at `60^(@)` to horizontal is 90m. What is the height reached when it is thrown at `30^(@)` to the horizontal:-

Options

A. 90m
B. 45m
C. 30m
D. 15m

Explanation

To solve the problem step by step, we will use the formula for maximum height reached by a projectile:

### Step 1: Understand the formula for maximum height
The maximum height \( h \) reached by a projectile is given by the formula:
\[ h = \frac{v_0^2 \sin^2 \theta}{2g} \]
where:
- \( v_0 \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 2: Calculate \( v_0^2 \) using the given height at \( 60^\circ \)
We know that the maximum height when the projectile is thrown at \( 60^\circ \) is \( 90 \, \text{m} \). Plugging in the values into the formula:
\[ 90 = \frac{v_0^2 \sin^2(60^\circ)}{2g} \]
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ 90 = \frac{v_0^2 \cdot \left(\frac{\sqrt{3}}{2}\right)^2}{2 \cdot 10} \]
\[ 90 = \frac{v_0^2 \cdot \frac{3}{4}}{20} \]
\[ 90 = \frac{3v_0^2}{80} \]

### Step 3: Solve for \( v_0^2 \)
Rearranging the equation:
\[ v_0^2 = \frac{90 \cdot 80}{3} \]
Calculating the right side:
\[ v_0^2 = \frac{7200}{3} = 2400 \]

### Step 4: Calculate the maximum height at \( 30^\circ \)
Now, we need to find the maximum height when the projectile is thrown at \( 30^\circ \):
\[ h = \frac{v_0^2 \sin^2(30^\circ)}{2g} \]
Substituting \( v_0^2 = 2400 \) and \( g = 10 \):
\[ h = \frac{2400 \cdot \left(\frac{1}{2}\right)^2}{2 \cdot 10} \]
\[ h = \frac{2400 \cdot \frac{1}{4}}{20} \]
\[ h = \frac{2400}{80} = 30 \]

### Final Answer
The height reached when the body is thrown at \( 30^\circ \) to the horizontal is \( 30 \, \text{m} \).

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