A body projected with velocity 30 `ms^(-1)` reaches its maximum height in 1.5 s. Its range is (g=`10ms^(-2)`)

Answer

To solve the problem, we need to determine the range of a projectile launched with an initial velocity of 30 m/s, reaching its maximum height in 1.5 seconds, with the acceleration due to gravity (g) being 10 m/s².

### Step-by-step Solution:

1. **Understanding the Components of Motion**:
The initial velocity (u) can be broken down into two components: horizontal (u cos θ) and vertical (u sin θ). At the maximum height, the vertical component of the velocity becomes zero.

2. **Using the Time to Reach Maximum Height**:
At maximum height, the vertical velocity (v_y) is 0. We can use the equation of motion:
\[ v_y = u_y - g t \]
Here, \( u_y = u \sin \theta \), \( v_y = 0 \), \( g = 10 \, \text{m/s}^2 \), and \( t = 1.5 \, \text{s} \). Thus, we have:
\[ 0 = u \sin \theta - g t \]
Substituting the values:
\[ 0 = u \sin \theta - 10 \times 1.5 \]
\[ u \sin \theta = 15 \]

3. **Finding the Angle θ**:
We know the initial velocity \( u = 30 \, \text{m/s} \). Thus:
\[ 30 \sin \theta = 15 \]
\[ \sin \theta = \frac{15}{30} = \frac{1}{2} \]
Therefore, \( \theta = 30^\circ \).

4. **Calculating the Range**:
The range (R) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
We need to calculate \( \sin 2\theta \):
\[ \sin 2\theta = \sin(2 \times 30^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \]
Now substituting the values into the range formula:
\[ R = \frac{(30)^2 \cdot \frac{\sqrt{3}}{2}}{10} \]
\[ R = \frac{900 \cdot \frac{\sqrt{3}}{2}}{10} \]
\[ R = \frac{900\sqrt{3}}{20} \]
\[ R = 45\sqrt{3} \, \text{m} \]

### Final Answer:
The range of the projectile is \( 45\sqrt{3} \, \text{m} \).

---