Question

Two balls are dropped from a tall building at the time interval of 3 second the 1st ball hits the ground rebound elastically and colloids death 2nd ball at a height of 78.75m. The height of building is. (G =10 ms-²)

Answer 1

Let t is the time taken to reach the height of building ‘ h ’.

when 1st ball hits the ground,
h + 78.75 = 0 + 1/2 × g × t²
h + 78.75 = 1/2 × 10 × t²
h = 5t²- 78.75 ..............(1)

when 1st ball collides with 2nd ball,
h - 78.75 = 1/2 × g × (t - 3)²
h - 78.75 = 1/2 × 10 × (t - 3)²
h - 78.75 = 5(t - 3)²
h = 78.75 + 5(t - 3)² .......(2)

from equations (1) and (2),
78.75 + 5(t - 3)² = 5t² - 78.75
78.75 + 78.75 = 30t - 45
157.5 + 45 = 30t
202.5 = 30t
t = 6.75 sec

h = 5(6.75)² - 78.75 = 149 m

hence, height of building = 149 m

Answer 2

HEY DEAR ... ✌️

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Here's , Your Answer ...

=) Assume , height of building (h) .

Now , When 1st ball hits the ground, 

h + 78.75 = 0 + 1/2 × g × t^2

h + 78.75 = 1/2 × 10 × t^2

h = 5t²- 78.75 ---------------(1)

Also , When 1st ball collides with 2nd ball ,

h - 78.75 = 1/2 × g × (t - 3)^2

h - 78.75 = 1/2 × 10 × (t - 3)^2

h - 78.75 = 5(t - 3)^2

h = 78.75 + 5(t - 3)^2 -----------(2)

From the above equations (1) and (2),

78.75 + 5(t - 3)^2 = 5t^2 - 78.75 

78.75 + 78.75 = 30t - 45 

157.5 + 45 = 30t 

202.5 = 30t 

t = 6.75 seconds

Now ,
h = 78.75 + 5(t - 3)^2
h = 5(6.75)^2 - 78.75
h = 149 m

Hence, the height of building is 149 m .

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HOPE , IT HELPS ... ✌️