A body projected obliquely with velocity `19.6ms^(-1)` has its kinetic energy at the maximum height equal to 3 times its potential energy there. Its position after t second of projection from the ground is (h = maximum height)

Options

A. `(h)/(2)`
B. `(h)/(4)`
C. `(h)/(3)`
D. h

Answer

To solve the problem step by step, we need to analyze the given information and apply the relevant physics concepts.

### Step 1: Understand the Problem
We are given that a body is projected obliquely with an initial velocity \( u = 19.6 \, \text{m/s} \). At maximum height, the kinetic energy (KE) is three times the potential energy (PE). We need to find the position of the body after \( t \) seconds of projection.

### Step 2: Write the Expressions for Kinetic and Potential Energy
1. The kinetic energy at maximum height is given by:
\[ KE = \frac{1}{2} m (u \cos \theta)^2 \]
2. The potential energy at maximum height is given by:
\[ PE = mgh = mg \left( \frac{u^2 \sin^2 \theta}{2g} \right) = \frac{mu^2 \sin^2 \theta}{2} \]

### Step 3: Set Up the Equation
According to the problem, we have:
\[ KE = 3 \times PE \]
Substituting the expressions for KE and PE, we get:
\[ \frac{1}{2} m (u \cos \theta)^2 = 3 \left( \frac{mu^2 \sin^2 \theta}{2} \right) \]
Cancelling \( m \) and \( \frac{1}{2} \) from both sides:
\[ (u \cos \theta)^2 = 3u^2 \sin^2 \theta \]

### Step 4: Simplify the Equation
Dividing both sides by \( u^2 \):
\[ \cos^2 \theta = 3 \sin^2 \theta \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \cos^2 \theta = 3(1 - \cos^2 \theta) \]
Let \( x = \cos^2 \theta \):
\[ x = 3(1 - x) \implies x + 3x = 3 \implies 4x = 3 \implies x = \frac{3}{4} \]
Thus, \( \cos^2 \theta = \frac{3}{4} \) and \( \sin^2 \theta = \frac{1}{4} \).

### Step 5: Find the Angle \( \theta \)
Taking the square root:
\[ \cos \theta = \frac{\sqrt{3}}{2}, \quad \sin \theta = \frac{1}{2} \]
This corresponds to \( \theta = 30^\circ \).

### Step 6: Calculate Maximum Height
Using the formula for maximum height:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
Substituting \( u = 19.6 \, \text{m/s} \), \( \sin^2 30^\circ = \frac{1}{4} \), and \( g = 9.8 \, \text{m/s}^2 \):
\[ H = \frac{(19.6)^2 \cdot \frac{1}{4}}{2 \cdot 9.8} = \frac{384.16 \cdot \frac{1}{4}}{19.6} = \frac{96.04}{19.6} = 4.9 \, \text{m} \]

### Step 7: Position After \( t \) Seconds
The height \( h \) after \( t \) seconds is given by:
\[ h = u \sin \theta \cdot t - \frac{1}{2} g t^2 \]
Substituting \( u = 19.6 \, \text{m/s} \) and \( \sin 30^\circ = \frac{1}{2} \):
\[ h = 19.6 \cdot \frac{1}{2} \cdot t - \frac{1}{2} \cdot 9.8 \cdot t^2 \]
\[ h = 9.8t - 4.9t^2 \]

### Step 8: Determine Position at \( t = 1 \) Second
Substituting \( t = 1 \):
\[ h = 9.8 \cdot 1 - 4.9 \cdot 1^2 = 9.8 - 4.9 = 4.9 \, \text{m} \]

### Conclusion
Since the maximum height \( H = 4.9 \, \text{m} \), the position after \( t = 1 \) second is \( \frac{H}{1} \).

### Final Answer
The position after \( t \) seconds of projection from the ground is \( H \).