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 \((R \times H) / T\) where:

- \(R\) is the range of the projectile,
- \(H\) is the maximum height attained,
- \(T\) is the time of flight.

Given the initial velocity \(\vec{u} = (5 \hat{i} + 10 \hat{j}) \, \text{m/s}\) and \(g = 10 \, \text{m/s}^2\), we can proceed with the following steps:

### Step 1: Calculate the Time of Flight (T)
The time of flight for a projectile can be calculated using the formula:

\[ T = \frac{2u_y}{g} \]

Where \(u_y\) is the initial vertical component of the velocity. Here, \(u_y = 10 \, \text{m/s}\).

Substituting the values:

\[ T = \frac{2 \times 10}{10} = 2 \, \text{s} \]

### Step 2: Calculate the Maximum Height (H)
The maximum height attained by the projectile can be calculated using the formula:

\[ H = \frac{u_y^2}{2g} \]

Substituting the values:

\[ H = \frac{10^2}{2 \times 10} = \frac{100}{20} = 5 \, \text{m} \]

### Step 3: Calculate the Range (R)
The range of the projectile can be calculated using the formula:

\[ R = u_x \times T \]

Where \(u_x\) is the initial horizontal component of the velocity. Here, \(u_x = 5 \, \text{m/s}\).

Substituting the values:

\[ R = 5 \times 2 = 10 \, \text{m} \]

### Step 4: Calculate \((R \times H) / T\)
Now we can substitute the values of \(R\), \(H\), and \(T\) into the expression:

\[ \frac{R \times H}{T} = \frac{10 \times 5}{2} \]

Calculating this gives:

\[ \frac{50}{2} = 25 \]

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

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