1.

A ball is thrown vertically downwards from a height of 20 m with an initial velocity yo2. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity vo is (Take g = 10 mS-2)

Answer»

V = v0 + G t
h = v0 t + g t²/2 = 20
(EVERYTHING MEASURED downwards)
Ek = (1/2) m v²
=> Ep = m g h = 196.2 m = (1/4) m v²
=> v ² = 784.8
=> v = 28.014 m/s
28.014 = v0 + 9.81 t
20 = v0 t + 4.905 t²
⇒ 20 = (28.014 - 9.81 t) t + 4.905 t²
= 28.014 t - 4.905 t²
⇒ 4.905 t² - 28.014 t + 20 = 0
This is a quadratic equation in t
t = (28.014 ± 19.809)/9.81
The SOLUTION with the - sign yields
t = 0.836 s ⇒ v0 = 19.82 m/s


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