Two balls are dropped from a tall building at a time
interval of 3 s. The first ball hits the ground, rebounds
elastically, and collides with second ball at a height
78.75 m. The height of the building is (g = 10 ms-2)
(1) 180 m
(2) 150 m
(3) 200 m
(4) 90 m
Answers
Answer:
hey mate here is your answer........,,
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
hope it's helpful for you..
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Answer:
h = 78.75 height at which balls collide
3 sec is time for ball1 to reach ground and bounce to h
1.5 sec is the time for ball 1 to travel each way to this point
h = 78.75 = v1 t + 1/2 g t^2 = 1.5 v1 + 1/2 * 10 + 2.25
v1 = 45 m/s speed of either ball at height
T = v1 / g = 45 / 10 = 4.5 sec time to fall to height h
t + T = 1.5 + 4.5 = 6 sec
H = 1/2 g t^2 = 5 * 36 = 180 m
Suggest you verify these values with equations such as
2 g h = v2^2 - v1^2 where v1 = 4.5 * 10 = 45
v2 = 6 * 10 = 60 which gives h as 78.75