Question

The initial and final velocities of a body projected vertically from the ground are \( 20 \, \text{ms}^{-1} \) and \( 18 \, \text{ms}^{-1} \) respectively. The maximum height reached by the body is (Acceleration due to gravity = \( 10 \, \text{ms}^{-2} \))

• A. 20 m
• B. 16.2 m
• C. 19 m
• D. 18.1 m

Hint:

If a body is projected upwards with velocity \(u\) and returns with velocity \(v\) under constant air resistance, the maximum height is given by: \[ H = \frac{u^2 + v^2}{4g} \]

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Since the body returns to the ground with a velocity (\(18 \, \text{ms}^{-1}\)) less than the initial velocity of projection (\(20 \, \text{ms}^{-1}\)), there is a loss of energy due to air resistance. We assume a constant retardation \(a\) due to air resistance. The effective acceleration magnitude during ascent is \((g+a)\) and during descent is \((g-a)\).
Step 2: Key Formula or Approach
Let \( H \) be the maximum height.
For ascent (final velocity at top is 0): \[ u^2 = 2(g+a)H \quad \Rightarrow \quad \frac{u^2}{2H} = g + a \quad \text{...(i)} \]
For descent (initial velocity at top is 0): \[ v^2 = 2(g-a)H \quad \Rightarrow \quad \frac{v^2}{2H} = g - a \quad \text{...(ii)} \]
Adding equations (i) and (ii) eliminates \( a \) and allows us to solve for maximum height \( H \) directly: \[ \frac{u^2}{2H} + \frac{v^2}{2H} = (g + a) + (g - a) \] \[ \frac{u^2 + v^2}{2H} = 2g \] \[ H = \frac{u^2 + v^2}{4g} \]
Step 3: Calculation
Given: \[ u = 20 \, \text{ms}^{-1}, \quad v = 18 \, \text{ms}^{-1}, \quad g = 10 \, \text{ms}^{-2} \] Substituting these values into the derived formula: \[ H = \frac{(20)^2 + (18)^2}{4 \times 10} \] \[ H = \frac{400 + 324}{40} \] \[ H = \frac{724}{40} \] \[ H = 18.1 \, \text{m} \] Final Answer:
The maximum height reached by the body is 18.1 m.