Question

For a body thrown vertically upwards from the ground, if the time interval between the two instants when the body is at a height of \(60\,\text{m}\) is \(4\,\text{s}\), then the maximum height reached by the body is \[ (\text{Acceleration due to gravity}=10\,\text{m s}^{-2}) \]

• A. \(150\,\text{m}\)
• B. \(90\,\text{m}\)
• C. \(120\,\text{m}\)
• D. \(80\,\text{m}\)

Hint:

For vertical projection, \[ \boxed{H=\frac{u^2}{2g}.} \] If a body reaches the same height twice, the difference between the two times can be obtained from the roots of the corresponding quadratic equation.

Answer 1

The Correct Option is D

Step 1: Form the equation of motion. The height after time \(t\) is \[ h=ut-\frac12gt^2. \] Given \[ h=60\,\text{m},\qquad g=10\,\text{m s}^{-2}, \] so \[ 60=ut-5t^2. \] or \[ 5t^2-ut+60=0. \]

Step 2: Use the time interval between the two roots. If the two times are \(t_1\) and \(t_2\), \[ t_2-t_1=4. \] For the quadratic, \[ t_2-t_1 = \frac{\sqrt{u^2-1200}}{5}. \] Hence, \[ \frac{\sqrt{u^2-1200}}5=4. \] Therefore, \[ u^2-1200=400, \] \[ u^2=1600, \] \[ u=40\,\text{m s}^{-1}. \]

Step 3: Find the maximum height. The maximum height is \[ H=\frac{u^2}{2g}. \] Substituting \[ u=40,\qquad g=10, \] \[ H = \frac{40^2}{2\times10} = \frac{1600}{20} = 80\,\text{m}. \] Hence, \[ \boxed{H=80\,\text{m}.} \] Therefore, the correct option is \(\boxed{(D)}\).