© All Rights reserved @ LearnWithDash
Step 1: Identify the Known Parameters
• Initial speed of the particle when projected upwards: $u = 16\,\text{m/s}$.
• Speed of the particle when it comes back to the same point: $v = 8\,\text{m/s}$.
• Let $h$ be the maximum height reached above the point of projection.
• Let $W$ be the magnitude of work done by air resistance during each phase (upward and downward).
• Acceleration due to gravity, $g \approx 10\,\text{m/s}^2$, acts downward.
Step 2: Apply Work-Energy Theorem for the Upward Journey
During the upward journey, the initial kinetic energy is partly converted into potential energy and partly lost due to the work done by air resistance $W$. Mathematically:
• Initial kinetic energy: $ \tfrac{1}{2} m (16)^2 $.
• Potential energy at maximum height: $ m g h $.
• Work done by air resistance (opposing upward motion): $ W $.
By the work-energy principle:
$ \tfrac{1}{2} m \times 16^2 = m g h + W $.
Step 3: Apply Work-Energy Theorem for the Downward Journey
On the way down from the maximum height to the original point, the particle loses potential energy $m g h$, and the same magnitude $W$ of energy is lost due to air resistance. Its final kinetic energy upon returning to the original point is $ \tfrac{1}{2} m \times 8^2 $.
Therefore:
$ \tfrac{1}{2} m \times 8^2 = m g h - W $.
Step 4: Solve the Equations to Find $h$
We have two equations:
(1) $ \tfrac{1}{2} m \times 16^2 = m g h + W \quad \Rightarrow \quad 128\,m = m g h + W$.
(2) $ \tfrac{1}{2} m \times 8^2 = m g h - W \quad \Rightarrow \quad 32\,m = m g h - W$.
Adding these two to eliminate $W$:
$128\,m + 32\,m = (m g h + W) + (m g h - W)$
$160\,m = 2\,m g h$
Simplifying gives:
$160 = 2\,g h \quad \Longrightarrow \quad g h = 80 \quad \Longrightarrow \quad h = \frac{80}{g}.$
Using $g \approx 10\,\text{m/s}^2$:
$h = \frac{80}{10} = 8\,\text{m}.$
Step 5: Final Answer
The maximum height attained by the particle is $8\,\text{m}$.
