Question

Derive an expression for the height attained by a freely falling body after 'n' number of ribbons from the floor.

Answer 1

You didn't say about coefficient of restitution , Let us consider that e is coefficient of restitution.

As you know, e=velocity of separation/velocity of approach

so,  in this case as ground is in rest so,

e = v/u , where u is velocity of body before striking the ground and v is the velocity of body after striking the ground.

every time it collides final velocity becomes initial thus power of 'e' increases.

for n collisions: final velocity(v)=(eⁿ)(initial velocity)

initial velocity = $\sqrt{2gh}$

so, (v)=(eⁿ )$\sqrt{2gh}$........(1)

thus, as it is uniform motion so, v² = 2gH.......(2)

From equations (1) and (2),

2gH=(e²ⁿ)2gh

thus,   H=(e²ⁿ)h

Answer 2

Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic and (c) non periodic motion? Give period for each case of periodic motion (ω is any positive constant) a. sign ωt - cos ωt b. sin² ωt c. 3 Cos (π/4 - ωt) d. cos ωt + cos 3ωt + cos 5ωt e. exp(-ω²t²) f. 1 + ωt + ω²t²