Question

Two balls are dropped from a tall building at time interval of 3s the 1st ball hits the ground rebounce, elastically and collides with 2nd ball at the height 78.75 m. The height of the building is (g=10 ms-¹)

Answer 1

say the total height of the building is h & say the time when it is colliding with the another ball in n, then,

$h+78.75=0+5n^{2} \\h-78.75=0+5(n-3)^{2} \\\\solve for h, ans coming is 149 meter.$

Answer 2

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