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A stone dropped from a building of height hh reaches the earth after tt seconds. From the same building, if two stones are thrown (one upwards and other downwards) with the same velocity uu and they reach the earth surface after t1t_1 and t2t_2 seconds respectively, then:

A

t=t1−t2t = t_1 - t_2

B

t=t1+t22t = \frac{t_1 + t_2}{2}

C

t=t1t2t = \sqrt{t_1 t_2}

D

t=t12t22t = t_1^2 t_2^2

Step-by-Step Solution

  1. Analyze the Dropped Stone: For a stone dropped from rest (u=0u=0), the height hh is given by h=12gt2h = \frac{1}{2}gt^2. Thus, t=2hgt = \sqrt{\frac{2h}{g}}.
  2. Analyze the Stone Thrown Upwards: Initial velocity is +u+u (upward), displacement is −h-h (downward). Using s=ut+12at2s = ut + \frac{1}{2}at^2: −h=ut1−12gt12  ⟹  h=−ut1+12gt12...(i)-h = u t_1 - \frac{1}{2}g t_1^2 \implies h = -u t_1 + \frac{1}{2}g t_1^2 \quad \text{...(i)}
  3. Analyze the Stone Thrown Downwards: Initial velocity is −u-u (downward), displacement is −h-h. Using the same equation: −h=−ut2−12gt22  ⟹  h=ut2+12gt22...(ii)-h = -u t_2 - \frac{1}{2}g t_2^2 \implies h = u t_2 + \frac{1}{2}g t_2^2 \quad \text{...(ii)}
  4. Solve for Relationship: From (i), u=12gt1−ht1u = \frac{1}{2}g t_1 - \frac{h}{t_1}. From (ii), u=ht2−12gt2u = \frac{h}{t_2} - \frac{1}{2}g t_2. Equating uu: 12gt1−ht1=ht2−12gt2\frac{1}{2}g t_1 - \frac{h}{t_1} = \frac{h}{t_2} - \frac{1}{2}g t_2 12g(t1+t2)=h(1t1+1t2)=ht1+t2t1t2\frac{1}{2}g (t_1 + t_2) = h \left( \frac{1}{t_1} + \frac{1}{t_2} \right) = h \frac{t_1 + t_2}{t_1 t_2} h=12gt1t2h = \frac{1}{2}g t_1 t_2 Comparing this with h=12gt2h = \frac{1}{2}gt^2, we get t2=t1t2t^2 = t_1 t_2 or t=t1t2t = \sqrt{t_1 t_2} .
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