A projectile is thrown with speed u making angle `theta` with horizontal at `t=0`. It just crosses the two points at equal height at time `t=1` s and `t=3` sec respectively. Calculate maximum height attained by it. `(g=10 m//s^(2))`

Explanation

To solve the problem step by step, we will analyze the motion of the projectile and use the information given about the times at which it crosses equal heights.

### Step 1: Understanding the motion of the projectile
The projectile is thrown with an initial speed \( u \) at an angle \( \theta \) with the horizontal. The vertical motion of the projectile is influenced by gravity, which acts downwards with an acceleration \( g = 10 \, \text{m/s}^2 \).

### Step 2: Setting up the equation for vertical motion
The vertical position \( y \) of the projectile at any time \( t \) can be expressed as:
\[ y(t) = u \sin \theta \cdot t - \frac{1}{2} g t^2 \]
At \( t = 1 \, \text{s} \) and \( t = 3 \, \text{s} \), the projectile is at the same height \( y \).

### Step 3: Writing the equations for the two times
At \( t = 1 \, \text{s} \):
\[ y(1) = u \sin \theta \cdot 1 - \frac{1}{2} g \cdot 1^2 = u \sin \theta - \frac{1}{2} \cdot 10 \cdot 1 = u \sin \theta - 5 \]
At \( t = 3 \, \text{s} \):
\[ y(3) = u \sin \theta \cdot 3 - \frac{1}{2} g \cdot 3^2 = u \sin \theta \cdot 3 - \frac{1}{2} \cdot 10 \cdot 9 = 3u \sin \theta - 45 \]

### Step 4: Setting the heights equal
Since both heights are equal:
\[ u \sin \theta - 5 = 3u \sin \theta - 45 \]

### Step 5: Rearranging the equation
Rearranging the equation gives:
\[ -5 + 45 = 3u \sin \theta - u \sin \theta \]
\[ 40 = 2u \sin \theta \]
\[ u \sin \theta = 20 \, \text{m/s} \]

### Step 6: Finding the maximum height
The maximum height \( H \) attained by the projectile is given by the formula:
\[ H = \frac{(u \sin \theta)^2}{2g} \]
Substituting \( u \sin \theta = 20 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ H = \frac{(20)^2}{2 \cdot 10} = \frac{400}{20} = 20 \, \text{m} \]

### Final Answer
The maximum height attained by the projectile is \( \boxed{20 \, \text{m}} \).