A ball is dropped from a height of 20m and renounce with the velocity which is 3/4 of d velocity with which it hits the ground. What is d tym interval btw the first and second bounces

1.73 seconds

Explanation:

The velocity the ball first hits the ground with is:

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (-10 m/s²) (-20 m)

v = -20 m/s

The velocity it rebounds with is 3/4 of that in the opposite direction, or 15 m/s.

The time it takes to return to the ground is:

Δx = v₀ t + ½ at²

0 = (15 m/s) t + ½ (-10 m/s²) t²

0 = t (15 − 5t²)

t = √3

t ≈ 1.73 seconds

like poles repel so keep pushing north to north and repels funny to see this

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s