a rubber ball falls on a floor from a height of 20m. Calculate velocity with which it strikes the ground. To what height will the ball rebound if it uses 40% of its energy on striking the ground.
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Given : Height from Which it has been Thrown = 20m
Usage Of KE after rebounce = 40%
Solution:
v² - u² = 2as
⇒ v² - o = 2 x 10 x 20
⇒ v = √400 = 20 m/s
KE = 1/2 x m x v² = 200 m Joules
40 % KE = (1/2) m x (v')² = [(200m) x 40]/100
on simplification we get v' = √160 m/s
again using
v² - u² = 2as
0 - u² = 2(-g)h'
⇒160 = 20h'
⇒h = 8 mts
∴height attained = after rebounce = 8 mts
Usage Of KE after rebounce = 40%
Solution:
v² - u² = 2as
⇒ v² - o = 2 x 10 x 20
⇒ v = √400 = 20 m/s
KE = 1/2 x m x v² = 200 m Joules
40 % KE = (1/2) m x (v')² = [(200m) x 40]/100
on simplification we get v' = √160 m/s
again using
v² - u² = 2as
0 - u² = 2(-g)h'
⇒160 = 20h'
⇒h = 8 mts
∴height attained = after rebounce = 8 mts
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