CBSE Class 11-science Answered

Let the height attained by the stone above the height of 5 mts is h and t₁ is the time taken to attain it. t₂ is the time taken to reach the ground from the highest point.  

For the motion from the height of 5 mts to highest point, using first equation of motion, we have

0 = u – gt₁

u = gt₁            ……………(1)

where u is the initial speed of the stone. Now using second equation of motion, we have

h = u t₁ - ½ g t₁²

Using equ. (1),

h = gt₁ t₁ - ½ g t₁² = ½ g t₁²       ………..(2)

For the motion from the highest point to the ground, using second equation of motion, we have

h + 5 = ½ g t₂²      (here the initial speed is zero)

eq. (2) – eq. (1)

5 = ½ g( t₂² - t₁²)

(t₂ + t₁)( t₂ - t₁) = 1

(t₂ - t₁) = 1 / 3.2

(t₂ + t₁) = 3.2

2 t₂ = 3.2 + 1/3.2 = (3.2 x 3.2 + 1) / 3.2

t₂ = 1.76

Therefore, h = (½) x 10 x (1.76)² – 5 = 10.49

Total distance covered = 2 h + 5 = 2 x 10.49 + 5 = 25.98 metre