Question

A body projected vertically up with an initial speed of $10\text{ ms}^{-1}$ reaches the point of projection after sometime with a speed of $8\text{ ms}^{-1}$. The maximum height reached by the body is (Acceleration due to gravity $= 10\text{ ms}^{-2}$)

• A. $5\text{ m}$
• B. $3.2\text{ m}$
• C. $4.1\text{ m}$
• D. $4.5\text{ m}$

Hint:

When speed decreases upon return to the same point, mechanical energy is lost due to non-conservative forces like air drag. Use Work-Energy principle separately for ascent and descent.

Answer 1

The Correct Option is C

Step 1: Analyze the Problem:
Since the body returns with a speed ($v = 8\text{ m/s}$) less than the projection speed ($u = 10\text{ m/s}$), there is a resistive force (air resistance) acting on the body. Assume a constant resistive force $f$. Let $H$ be the maximum height.
Step 2: Work-Energy Theorem:
During the upward journey, work is done against gravity and air resistance. Initial KE = Work done against gravity + Work done against resistance. \[ \frac{1}{2}mu^2 = (mg + f)H \quad \dots(1) \] During the downward journey, potential energy is converted to kinetic energy and work against resistance. \[ (mg - f)H = \frac{1}{2}mv^2 \quad \dots(2) \]
Step 3: Solve for Resistance Force $f$:
Divide (1) by (2): \[ \frac{mg+f}{mg-f} = \frac{u^2}{v^2} = \frac{10^2}{8^2} = \frac{100}{64} = \frac{25}{16} \] \[ 16(mg+f) = 25(mg-f) \] \[ 16mg + 16f = 25mg - 25f \] \[ 41f = 9mg \implies f = \frac{9}{41}mg \]
Step 4: Calculate Maximum Height $H$:
Substitute $f$ back into equation (1): \[ \left(mg + \frac{9}{41}mg\right)H = \frac{1}{2}m(100) \] \[ mg\left(1 + \frac{9}{41}\right)H = 50m \] \[ g\left(\frac{50}{41}\right)H = 50 \] \[ 10 \cdot \frac{50}{41} H = 50 \] \[ \frac{10}{41} H = 1 \implies H = 4.1\text{ m} \]