A projectile is projected from ground with an initial velocity `(5hati+10hatj)m//s` If the range of projectile is R m, time of flight is Ts and maximum height attained above ground is Hm, find the numerical value of `(RxxH)/T`. [Take `g=10m//s^2`]

Explanation

To solve the problem, we need to find the numerical value of \((R \times H) / T\) given the initial velocity of the projectile as \((5 \hat{i} + 10 \hat{j}) \, \text{m/s}\) and \(g = 10 \, \text{m/s}^2\).

### Step 1: Determine the initial velocity components
The initial velocity \( \mathbf{u} \) can be broken down into its components:
- \( u_x = 5 \, \text{m/s} \) (horizontal component)
- \( u_y = 10 \, \text{m/s} \) (vertical component)

### Step 2: Calculate the time of flight \( T \)
The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2 u_y}{g} \]
Substituting the values:
\[ T = \frac{2 \times 10}{10} = 2 \, \text{s} \]

### Step 3: Calculate the range \( R \)
The range \( R \) of the projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
First, we need to find \( \theta \):
\[ \tan(\theta) = \frac{u_y}{u_x} = \frac{10}{5} = 2 \implies \theta = \tan^{-1}(2) \]
Now, calculate \( \sin(2\theta) \):
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
Using the values:
\[ \sin(\theta) = \frac{u_y}{u} = \frac{10}{\sqrt{5^2 + 10^2}} = \frac{10}{\sqrt{125}} = \frac{10}{5\sqrt{5}} = \frac{2}{\sqrt{5}} \]
\[ \cos(\theta) = \frac{u_x}{u} = \frac{5}{\sqrt{125}} = \frac{5}{5\sqrt{5}} = \frac{1}{\sqrt{5}} \]
Thus,
\[ \sin(2\theta) = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5} \]
Now substituting back into the range formula:
\[ R = \frac{(5\sqrt{5})^2 \cdot \frac{4}{5}}{10} = \frac{125 \cdot \frac{4}{5}}{10} = \frac{100}{10} = 10 \, \text{m} \]

### Step 4: Calculate the maximum height \( H \)
The maximum height \( H \) is given by:
\[ H = \frac{u_y^2}{2g} \]
Substituting the values:
\[ H = \frac{10^2}{2 \cdot 10} = \frac{100}{20} = 5 \, \text{m} \]

### Step 5: Calculate \( \frac{R \times H}{T} \)
Now we can find the required value:
\[ \frac{R \times H}{T} = \frac{10 \times 5}{2} = \frac{50}{2} = 25 \]

### Final Answer
The numerical value of \((R \times H) / T\) is \( 25 \).