Question

A body thrown vertically upwards with certain velocity from the ground reaches a maximum height \( H \). The ratio of the times at which the body is at a height of \( \frac{H}{2} \) is:

• A. \( 3:2 \)
• B. \( \sqrt{3} : \sqrt{2} \)
• C. \( (\sqrt{3} - 1) : (\sqrt{3} + 1) \)
• D. \(\mathbf{(\sqrt{2} -1) : (\sqrt{2} +1)}\)
  \

Hint:

When solving projectile motion problems: - Use energy conservation or kinematics equations. - For time ratio at different heights, set up the quadratic equation using vertical motion equations. - Solve for \( t_1 \) and \( t_2 \) and express them as a ratio.

Answer 1

The Correct Option is D

Step 1: Using the kinematic equation for vertical motion The equation of motion for vertical displacement: \[ v^2 = u^2 - 2g s \] At maximum height \( H \), the final velocity is zero: \[ 0 = u^2 - 2gH \] Solving for \( u^2 \): \[ u^2 = 2gH \] Now, considering the time to reach height \( \frac{H}{2} \), we use: \[ H/2 = u t - \frac{1}{2} g t^2 \] Substituting \( u = \sqrt{2gH} \): \[ \frac{H}{2} = \sqrt{2gH} t - \frac{1}{2} g t^2 \] Rearranging: \[ g t^2 - 2\sqrt{2gH} t + H = 0 \] Step 2: Solving the quadratic equation Dividing throughout by \( g \): \[ t^2 - \frac{2\sqrt{2H}}{g} t + \frac{H}{g} = 0 \] Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ t = \frac{\frac{2\sqrt{2H}}{g} \pm \sqrt{\left( \frac{2\sqrt{2H}}{g} \right)^2 - 4 \times \frac{H}{g}}}{2} \] \[ t = \frac{\frac{2\sqrt{2H}}{g} \pm \sqrt{\frac{8H}{g} - \frac{4H}{g}}}{2} \] \[ t = \frac{\frac{2\sqrt{2H}}{g} \pm \sqrt{\frac{4H}{g}}}{2} \] \[ t = \frac{\frac{2\sqrt{2H}}{g} \pm \frac{2\sqrt{H}}{g}}{2} \] \[ t = \frac{2\sqrt{H}(\sqrt{2} \pm 1)}{2g} \] \[ t = \frac{\sqrt{H}(\sqrt{2} \pm 1)}{g} \] Step 3: Finding the ratio of the times Since the two possible values of \( t \) are: \[ t_1 = \frac{\sqrt{H}(\sqrt{2} -1)}{g}, \quad t_2 = \frac{\sqrt{H}(\sqrt{2} +1)}{g} \] The required ratio is: \[ \frac{t_1}{t_2} = (\sqrt{2} -1) : (\sqrt{2} +1) \] Step 4: Verifying the correct option Comparing with the given options, the correct answer is: \[ \mathbf{(\sqrt{2} -1) : (\sqrt{2} +1)} \]