Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
= e ( coefficient of restitution ) = 
and

h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that


So on

= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
The answer is most likely A
Answer:
2 m/s²
Explanation:
the equations of motion are
S= ut +½at²
v² = u²+ 2as
v = u + at
s = (u+v)/2 × t
From the parameters given
u = 0m/s this is because it starts from rest
Distance (s) = 9m
Time (t) = 3s
Based on this the first equation would be used
s = ut + ½at²
Input values
9 = 0×3 + ½ × a x 3²
9 = 0 + 9a/2
9 = 4.5a
Divide both sides by 4.5
a = 9 / 4.5 m/s²
a = 2 m/s²
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Answer:
Part a)
When spring compressed by 2 cm
H = 1.47 m
Part b)
When spring is compressed by 4 cm
H = 5.94 m
Explanation:
Part a)
As we know that the spring is compressed and released
so here spring potential energy is converted into gravitational potential energy at its maximum height
So we will have


so we have

Part b)
Similarly when spring is compressed by 4 cm
then we have


so we have

a tendency of a body to do nothing or to remain unchanged.