# A ball is allowed to fall down with initial speed v from a height of 10 m. It loses 50% kinetic energy…

Q: A ball is allowed to fall down with initial speed u from a height of 10 m. It loses 50% kinetic energy after striking the floor. It reaches to the same height after collision. What is the value of u ?

(a) 28 m/s

(b) 7 m/s

(c) 14 m/s

(d) it is never possible

Ans: (c)

Sol:

To reach same height of 10 m after collision,

Velocity After collision $\displaystyle = \sqrt{2 g h}$

$\displaystyle = \sqrt{2 \times 10 \times 10} = 10\sqrt{2} m/s$

As energy lost in collision is 50% , velocity becomes v⁄√2

∴ Velocity of ball on striking,

v = (10 √2) √2 = 20 m/s

Applying $\displaystyle v^2 = u^2 + 2 a S$

$\displaystyle 20^2 = u^2 + 2 \times 10 \times 10$

$\displaystyle u^2 = 20^2 – 200 = 200$

$\displaystyle u = \sqrt{200} = 10\sqrt{2}$

u = 14.14 m/s